Main content

# Curve sketching with calculus: logarithm

Sal sketches a graph of f(x)=ln(x⁴+27) including extremum and inflection points. Created by Sal Khan.

## Want to join the conversation?

- Instead of saying ". . just less/greater than. . ."

Wouldn't it just be "just in that sign direction until the critical value?"

So at13:24, Instead of saying, "So when x is slightly less than zero. . " Would it be the same as saying, "When x is between 0 and -3. . . "?

(If there are no more critical points between the value and +/- infinity, then you would just say, "When x is between 0 and negative infinity. . "_

If not, I would like to know why.

Thanks.(19 votes)- You are correct. I think Sal was just trying to stress the meaning of the inequality.

In general there are many ways to say the same thing. Some ways are more formal, some are more standard, some are easier to understand. A good mathematician is always thinking about the best way to say something just like you are.(11 votes)

- I don't completely understand why x = 0 is a minimum point. Sal's logic does make sense (~15:14) that since f"x is positive at both sides of 0, it is concave up all around and x = 0 is a minimum, but in the last video Sal said that in order to verify that a critical point is a min/max you should take the second derivative at that point. If the second derivative is positive it's a min, if it's negative it's a maximum and if it's zero it is neither but may be a potential inflection point. But, if you take the second derivative of zero, in this case, it is zero, so with that logic shouldn't it not be a minimum?(11 votes)
- Try taking the second derivative of the
**function**, and then evaluate it at the critical point.(0 votes)

- If you found the third derivative of a function and you plug in your inflection point candidates into it and you get 0 for one of them, does that immediately tell you that that specific candidate is NOT an inflection point?(4 votes)
- No. If the third derivative is 0, the test in inconclusive. But if the third derivative exists and is non-zero at a point where the second-derivative is zero, then that point is an inflection point. If the third derivative is 0, you cannot tell one way or the other.(5 votes)

- Since we know that from -infinity to -3 and from 3 to infinity, the graph is going to be concave downwards, how did we know that the graph doesn't become steeper and steeper until it's basically a vertical line? This may be a little bit confusing wording, but I thought the graph would take a more "m" shape where it would continue sloping down faster and faster as we approached infinity from the positive/negative directions.

I graphed it and it seems to slow its negative acceleration down until it almost looks like a horizontal line. I'm just a bit confused about how Sal knew this when he graphed it.(4 votes)- That's a great question. If the graph does become a vertical line, then the first derivative wouldn't be defined. So as long as the first derivative function is defined as x approaches plus or minus infinity, then you can be assured the function doesn't become a vertical line.(2 votes)

- If you have a set of possible inflection points, why pick numbers slightly above/below each, instead of any point in between? For example, if you have possible inflections at 0 and 3, why check for positive/negative at both 0 plus a little and 3 minus a little, instead of just checking f''(1) and using it for both?(3 votes)
- You do only need to check f ''(1). (That is, as long as your second derivative is continuous; if you have a discontinuity somewhere you have to consider that as a point where the concavity could change)(2 votes)

- in the AP calculus AB exam, would there be a question regarding sketching? It seems time consuming..(3 votes)
- This is very confusing.

How did he put everything together?

I thought I had to find critical point with the first derivative and apply the "first derivative test" to find critical points and global/local max/min, and then use the "second derivative test" searching for candidate (new critical points) to be inflection points and to find concavity.

How did he find the min/max just from the 2nd derivative? and how did he manage to reuse the same critical points?

Please help! thanks in advance!(3 votes) - I'm confused as to why when using the product rule [f'(x)g(x) + f(x)g'(x)] to calculate the second derivative. You first define the f(x) to be 4x^3 and g(x) to be (x^4 + 27)^-1. Where then, when calculating the second part of the product rule equation [f(x)g'(x)], does he get the second 4x^3? If f(x)=4x^3 and g'(x)= -1(x^4+27)^-2. Why does he multiply 4x^3 twice and then g'(x). Wouldn't that be f(x)f(x)g'(x)?(2 votes)
- Pay attention to the colours used. You get the second 4x^3 from the derivative of the inside of (x^4+27)^-1. g'(x) has to be -1(x^4+27)^-2*4x^3. Say, you have a function h(x)^-1. The derivative is -1*h(x)^-2*h'(x).(2 votes)

- I'm so confused.. at about11:00Sal decided to factor out 27·12x^2 - 4x^6 = 0 to 4x^2(27·3-x^4) = 0 .. When I was doing it on my own I multiplied the constants and got 324x^2-4x^6 = 0, factoring out to 4x^2(324-x^4) = 0... Is this an example of where BEDMAS is really important?(2 votes)
- You forgot to divide the 324 in your final equation by 4. :-)(1 vote)

- Is it possible to figure out if 0 is an inflection point by writting a limit instead of just taking really close to zero numbers?(2 votes)
- No. That won't work for an inflection point. There are three requirements for an inflection point.

There is an inflection point at x=c if and only if:

1. f(x) is continuous at x=c

AND

2. f''(c) = 0

AND

3. f''(c-ε) has a different sign from f''(c+ε)

This test always works.

A few (not most) inflection points can be found by the first derivative test:

If f(x) is continuous at x=c ,

AND

If f'(c) = 0

AND

If f'(c-ε) has the same sign as f'(c+ε)

THEN

f(x) has an inflection point at x=c

In other words, if the first derivative is 0 at some point but it is not a max or min, then it is an inflection point. However, most inflection points cannot be found this way because they will be located at some point where the first derivative is not 0.(1 vote)

## Video transcript

Let's say we've got the
function, f of x is equal to the natural log of x
to the fourth plus 27. And all we want to do is take
its first and second derivatives, and use as much of
our techniques as we have at our disposal to attempt
to graph it without a graphing calculator. If we have time, I'll take out
the graphing calculator and see if our answer matches up. So a good place to start
is to take the first derivative of this. So let me do that over here. So the derivative of f. Well, you take the derivative
of the inside, so take the derivative of that right there,
which is 4x to the third, and then multiply it times the
derivative of the outside, with respect to the inside. So the derivative of the
natural log of x is 1over x. So the derivative of this whole
thing with respect to this inside expression is going
to be, so times 1 over x to the fourth plus 27. If you found that confusing,
you might want to rewatch the chain rule videos. But that's the first
derivative of our function. I could rewrite this, this is
equal to 4x to the third over x to the fourth, plus 27. Or I could write it as 4x to
the third times x to the fourth, plus 27 to
the negative 1. All three of these
expressions are equivalent. I'm just the writing, I
multiplied it out, or I could write this as a negative
exponent, or I could write this as a fraction, with
this in the denominator. They're all the equivalent. So that's our first derivative. Let's do our second derivative. Our second derivative,
this looks like it'll get a little bit hairier. So our second derivative is
the derivative of this. So it's equal to, we can now
use the product rule, is the derivative of this first
expression, times the second expression. So the derivative of this first
expression, 3 times 4is 12. 12x squared, right, we just
decrement the 3 by 1, times the second expression, times x to
the fourth plus 27 to the minus 1, and then to that, we want to
add just the first expression, not its derivative, so just 4x
to the third, times the derivative of the
second expression. And the derivative of the
second expression, we could take the derivative of the
inside, which is just 4x to the third, the derivative of 27 is
just 0, so times 4x to the third, times the derivative
of this whole thing with respect to the inside. So times, so you take this
exponent, put out front, so times minus 1, times this whole
thing, x to the fourth plus 27 to the, we decrement this
by one more, so minus two. So let's see if I can simplify
this expression a little bit. So this is equal to, so this
right here is equal to 12x squared over this thing, x to
the fourth plus 27, and then, let's see, if we multiply we're
going to have a minus here, so it's a minus, you multiply
these two guys, 4 times 4is 16, 16x to the third times x to the
third is x to the sixth, over this thing squared. Over x to the fourth
plus 27 squared. That's just another way to
rewrite that expression right there, right? To the minus 2, you just put in
the denominator and make it into a positive 2 in
the denominator. Same thing. Now, if you've seen these
problems in the past, we always want to set these
things equal to 0. We want to solve
for x equals 0. So it'll be useful to have this
expressed as just one fraction, instead of the difference or
the sum of two fractions. So what we can do, is we could
have a common denominator. So we could multiply both the
numerator and denominator of this expression by x to the
fourth plus 27, and what do we get? So this is equal to, so if we
multiply this first expression, times x to the fourth plus 27,
we get 12x squared, times x to the fourth, plus 27. And then in the denominator,
you have x to the fourth plus 27 squared. All I did, I multiplied this
numerator and this denominator by x to the the fourth plus 27. I didn't change it. And then we have
that second term. Minus 16 x to the sixth over x
to the fourth plus 27 squared. The whole reason why did that? Now I have a common
denominator, now I can just add the numerators. So this is going to be
equal to, let's see. The denominator, we know what
the denominator is, it is x to the fourth plus 27 squared. That's our denominator. And then we can
multiply this out. This is 12x squared
times x to the fourth. That's 12x to the sixth,
plus 27 times 12. I don't even feel like
multiplying 27 times 12,so I'll just write that out. So plus 27 times 12x squared, I
just multiplied the 12x squared times the 27, and then minus
16x to the sixth minus 16x to the sixth. And this simplifies to,
let's see if I can simplify this even further. 7x to the sixth here,
x to the sixth here. So this is equal to,
do this in pink. This is equal to the 27 times
12x squared, I don't feel like figuring that out right now,
times 12x squared, and then you have minus 16x to the sixth
and plus 12x to the sixth. So you add those two,
you get minus 4. 12 minus 15is minus 4, x to the
sixth, all of that over x to the fourth plus 27
plus 27 squared. And that is our
second derivative. Now, we've done all of the
derivatives, and this was actually a pretty
hairy problem. And now we can solve for when
the first and the second derivatives equal 0, and we'll
have our candidate, well, we'll know our critical points, and
then we'll have our candidate inflection points, and
see if we can make any headway from there. So first, let's see where our
first derivative is equal to 0, and get our critical points. Or at least maybe, also
maybe, where it's undefined. So this is equal to 0. If we want to set, if the only
place that this can equal to 0 is if this numerator
is equal to 0. This denominator, actually, if
we are assuming we're dealing with real numbers, this term
right here is always going to be greater than or equal to 0
for any value of x, because it's an even exponent. So this thing can never equals
0, right, because you're adding 27 to something
that's non-negative. So this will never equal
0, so this will also never be undefined. So there's no undefined
critical points here, but we could set the numerator
equal to 0 pretty easily. If we wanted to set this equal
to 0, we just say 4x to the third is equal to 0, and we
know what x-value will make that equal to 0, x has
to be equal to 0. 4 times something to the
third is equal to 0, that something has to be 0. x to the third has to
be 0, x has to be 0. So we can write, f prime
of 0 is equal to 0. So 0 is a critical point. 0 is a critical point. The slope at 0 is 0. We don't know if it's a
maximum or a minimum, or an inflection point yet. We'll explore it a
little bit more. And actually, just so
we get the coordinate, what's the coordinate? The coordinate x is 0, and then
y is the natural log-- if x is 0, this just turns out, it's
the a natural log of 27. Let me figure out what that is,
I'll get the calculator out. I said I wouldn't use a
graphing calculator, but I can use a regular calculator. So27, if I were to take the
natural log of that, for our purposes let's
just call it 3.3. We're just trying to get the
general shape of the graph. So 3.3. Well, we could just say
2.9 and it kept going. So this is a critical
point right here. The slope is 0 here. Slope is equal to 0
at x is equal to 0. So this is one thing
we want to block off. And let's see if we can
find any candidate inflection points. And remember, candidate
inflection points are where the second derivative equals 0. Now if the second derivative
equals 0, that doesn't tell us that those are definitely
inflection points. Let me make this very clear. If, let me do it
in a new color. If x is inflection, then the
second derivative at x is going to be equal to 0. Because you're having
a change concavity. You have a change in the slope,
goes from either increasing to decreasing or from
decreasing to increasing. But if the derivative is equal
to 0, the second derivative is equal to 0, you cannot assume
that is an inflection point. So what we're going to do is,
we're going to find all of the point at which this is true,
and then see if we actually do have a sign change in the
second derivative of that point, and only if you have a
sign change, then you can say it's an inflection point. So let's see if we can do that. So just because a second
derivative is 0, that by itself does not tell you
it's an inflection point. It has to have a second
derivative of 0, and when you go above or below that x, the
second derivative has to actually change signs. Only then. So we can say, if f prime
changes signs around x, then we can say that
x is an inflection. And if it's changing signs
around x, then it's definitely going to be 0 right at x, but
you have to actually see that if it's negative before x, has
to be positive after x,or if it's positive before x, has
to be negative after x. So let's test that out. So the first thing we
need to do is find these candidate points. Remember, the candidate
points are where the second derivative is equal to 0. We're going to find those
points, and then see if this is true, that the sign
actually changes. We want to find where this
thing over here is equal to 0. And once again, for this to
be equal to 0, the numerator has to be equal to 0. This denominator can never be
equal to 0 if we're dealing with real numbers, which I
think is a fair assumption. So let's see where this our
numerator can be equal to 0 for the second derivative. So let's set the numerator
of the second derivative. 27 times 12x squared minus 4x
to the sixth is equal to 0. Remember, that's just
the numerator of our second derivative. Any x that makes the
numerator 0 is making the second derivative 0. So let's factor
out a 4x squared. So 4x squared. Now we'll have 27 times, if we
factor 4 out of the 12, we'll just get a 3, and we factored
out the x squared, minus, we factored out the 4, we factored
out an x squared, so we have x to the fourth is equal to 0. So the x's that will make this
equal to 0 will satisfy either, I'll switch colors, either 4x
squared is equal to 0, or, now 27 times 3, I can do
that in my head. That's 81. 20 times 3 is 60, 7 times
3 is 21, 60 plus 21is 81. Or 81 minus x to the
fourth is equal to 0. Any x that satisfies either
of these will make this entire expression equal 0. Because if this thing is
0, the whole thing is going to be equal to 0. If this thing is 0, the
whole thing is going to be equal to 0. Let me be clear, this
is 81 right there. So let's solve this. This is going to be 0 when
x is equal to 0, itself. This is going to be equal
to 0 when x, let's see. If we add x to the fourth to
both sides, you get x to the fourth is equal to 81. If we take the square root of
both sides of this, you get x squared is equal to 9, or so
you get x is plus or minus 3. x is equal to plus
or minus three. So these are our candidate
inflection points, x is equal to 0, x is equal to plus 3,
or x is equal to minus 3. So what we have to do now, is
to see whether the second derivative changes signs around
these points in order to be able to label them
inflection points. So what happens when x
is slightly below 0? So let's take the situation,
let's do all the scenarios. What happens when x
is slightly below 0? Not all of them, necessarily,
but if x is like 0.1. What is the second derivative
going to be doing? If x is 0.1, or if x is minus
0.1, this term right here is going to be positive, and
then this is going to be 81 minus 0.1 to the fourth. So that's going to be a
very small number, right? So it's going to be some
positive number times 81 minus a small number. So it's going to be
a positive number. So when x is less than 0, or
just slightly less than 0, our second derivative is positive. Now what happens when
x is slightly larger? When I write this notation, I
want to be careful, I mean, really, just right below 0. Now when x is right
above 0, what happens? Let's say x was 0.01,
or 0.1, positive 0.1. Well, it's going to
be the same thing. Because in both cases,
we're squaring, and we're taking the fourth. So you're kind of losing
your sign information. So if x is 0.1, this thing
is going to be a small positive number. You're going to be subtracting
a very small positive number from 81, but 81 minus a small
number is still going to be positive. So you're going to positive
times a positive, so your second derivative is still
going to be greater than 0. So something interesting here. f at your second derivative is
0 when x is equal to 0, but it is a not an inflection point. Because notice, the concavity
did not change around 0. Our second derivative is
positive as we approach 0 from the left, and it's positive as
we approach 0 from the right. So in general, at 0, we're
always, as we're near 0 from either direction, we're going
to be concave upwards. So the fact that 0 is a
critical point, and that we're always concave upward, as we
approach 0 from either side, this tells us that this
is a minimum point. Because we're concave
upwards all around 0. So 0 is not an
inflection point. Let's see if positive
and negative 3 are inflection points. And if you study this
equation, let me write our-- and actually, I
just want to be clear. I've just been using
the numerator of the second derivative. The whole second derivative is
this thing right here, but I've been ignoring the denominator
because the denominator is always positive. So if we're trying to
understand whether things are positive or negative, we just
really have to determine whether the numerator is
positive or negative. Because this expression right
there is always positive. It's something to
the second power. So let's test whether we have a
change in concavity around x is equal to positive
or negative 3. So remember, the numerator of
our, let me just rewrite our second derivative, just so
you see it here. f prime prime of x. The numerator is this
thing right here. It's 4x squared times 81
minus x to the fourth. and the denominator
was up here, x to the fourth plus 27 squared. That was our second derivative. Let's see if this changes signs
around positive or negative 3. And actually, we should get the
same answer, because regardless of whether we put positive
or negative 3 here. you lose all your sign
information because you're taking it to the fourth
power, you're taking it to the second power. And obviously, anything to the
fourth power is always going to be positive, anything to the
second power is always going to be negative. So when we do our test, if it's
true for positive 3, it's probably going to be true
for negative 3 as well. But let's just try it out. So when x is just a little bit
less than positive 3, what's the sign of f prime prime of x? So it's going to be 4 times
9, or it's going to be 4 times a positive number. It might be like 2.999,
but this is still going to be positive. So this is going to be positive
when x is approaching 3, and then this is going to be, well,
if x is 3, this is 0, so x is a little bit less than 3. If x is a little bit less than
3, if it's like 2.9999, this number is going to be less
than 81, so this is also going to be positive. And of course, the denominator
is always positive. So as x is less than 3, is
approaching from the left, we are concave upwards. This thing's going
to be a positive. Then f prime prime
is greater than 0. We are upwards,
concave upwards. When x is just larger than
3, what's going to happen? Well, this first term is
still going to be positive. But if x is just larger than 3,
x to the fourth is going to be just larger than 81, and so
this second term is going to be negative in that situation. Let me do it ina new color. It's going to be negative
when x is larger than 3. Because this is going
to be larger than 81. So if this is negative and this
is positive, then the whole thing is going to be negative,
because this denominator is still going to be positive. So then f prime prime is going
to be less than 0, so we're going to be concave downwards. One last one. What happens when x is just
a greater than minus 3? So just being greater
than minus 3, that's like minus 2.99999. So when you take minus 2.99
square it, you're going to get a positive number, so this
is going to be positive. And if you take minus 2.99 to
the fourth, that's going to be a little bit less
than 81, right? Because 2.99 to the fourth is a
little bit less than 81, so this is still going
to be positive. So you have a positive times a
positive divided by a positive, so you're going to be concave
upwards, because your second derivative is going to
be greater than 0. Concave upwards. And then finally, when x is
just, just less than negative 3, remember, when I write this
down, I don't mean for all x's larger than negative 3,or all
x's smaller than negative 3. There's actually no, well, I
can't think of the notation that would say just, as we just
approach three in this case, from the left, but what happens
if we just go to minus 3.11? Or 3.01, I guess is a
better one, or 3.1? Well, this term right here
is going to be positive. But if we take minus 3.1 to the
fourth, that's going to be larger than positive 81, right? The sign will become positive,
it'll be larger than 81, so this'll become negative. So in that case as well, we'll
have a positive times a negative divided by a positive,
so then our second derivative is going to be negative. And so we're going
to be downwards. So I think we're ready to plot.
so first of all, is x plus or minus 3 inflection points? Sure! As we approach x is equal to 3
from the left, we are concave upwards, and then as we cross
3, the second derivative is 0. The second derivative's
0, I lost it up here. The second derivative is 0. And then, as we go to the
right of 3, we become concave downwards. So we got our sign change
in the second derivative. So x is equal to 3. So 3 is definitely an
inflection point, and the same argument could be
made for negative 3. We switch signs as we cross 3. So these definitely are
inflection points. Just so we get the exact
coordinates, let's figure out what f of 3 is, or f of
positive and negative 3. And then we're ready to graph. So first of all, we know that
f, we know that the point 0, 3.29, that this was a minimum. Because 0 was a critical
point, the slope is 0 there, and because it's concave
upwards all around 0. So 0 is definitely not
an inflection point. And then we know that the
points positive 3 and minus 3 are inflection points, and in
order to figure out their y-coordinates, we can
just evaluate them. So they're actually going to
have the same y-coordinates, because if you put a minus3 or
positive 3 and take it to the fourth power, you're going
to get the same thing. Let's figure out what they are. So if we take 3 to the fourth
power, that's what, 81. 81 plus 27 is equal to 108,
and then we want to take the natural log of it. Let's just say 4.7, just
to get a rough idea. That's4.7. And that's true of whether we
do positive or negative 3, because we took to
the fourth power. So it's 4.7, 4.7. These are both
inflection points. And we should be
ready to graph it! Let's graph it. All right. Let me draw my axis,
just like that. And this is my y-axis, this
is my x-axis, this is y. You can even call it the f
of x axis, if you like. This is x. And so the point 0, 3.29. Let's say this is 1, 2, 3,
4, 5, to the point 0, 3.29. That's 0, 1, 2, 3, a little bit
above 3, it's right there. That's the minimum point. And then we're concave. The slope is 0 right there, we
figured that out, because the first derivative was 0 there. So it's a critical
point, and it's concave upwards around there. So that told us we arepoint at
a minimum point, right there. And then at positive 3. So 1, 2, 3. At positive 3, 4.7. So 4.7 will look
something like that. We have an inflection point. Before that, we're concave
upwards, and then after that we're concave downwards. So it looks something
like this. So we're concave upwards
up with up to that point. Maybe, actually, you should,
let me ignore that yellow thing I drew before. Let me get rid of that. Let me draw it like 1, 2, 3. 3, 4.7 looks like that, and
minus 3, 4.7, 1, 2, 3, 4.7 looks like that. So we know at 0, we are slope
of 0 and we're concave upwards, so we look like this. We're concave upwards,
until x is equal to 3. And at x is equal to 3, we
become concave downwards, and we go, let me try my best to
draw it well, and we go off like that. And then we're concave upwards
around 0, until we get, we're concave upwards as long as x is
greater than minus 3, and then at minus 3 we become
concave downwards again. Maybe I should do
it in that color. This concave downwards right
here, that's this, right here. That's that, right there. And this concave downwards,
right here-- sorry, I meant to do it in the red color--
this concave downwards right here, is this, right there. And then the concave upwards
around 0 is right there. You could even imagine, this
concave upwards that we measured, that's this, concave
upwards, and then this concave upwards is that. And then around 0,
we're always upwards. So this is my sense of what the
graph will look like and maybe it'll just you know it turns
into well you could think about what it does is x approaches
positive or negative infinity, some of the terms, well,
I won't go into that. But let's test whether we've
grafted correctly using a graphing calculator. So let me get out my TI-85,
trusty TI-85, and let's graph this sucker. All right, press graph. y equals the natural log of
x to the fourth plus 27. All right, I want to
hit that graph there. So I do second, graph. And let's cross our fingers. It looks pretty good! It looks almost exactly
like what we drew. So I think our I think our
mathematics was correct. This was actually
very satisfying. So hopefully you appreciate the
usefulness of inflection points, and second derivative,
and first derivative, in graphing some of
these functions.