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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 5

Lesson 9: Sketching curves of functions and their derivatives

# 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 at , 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. • 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.
• I don't completely understand why x = 0 is a minimum point. Sal's logic does make sense (~) 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? • The local max / min (also called "extrema") occur at the bottom or top of a curve. At a max, the curve stops increasing and changes direction to start decreasing (the slope switches from positive to negative). Thus the slope at that exact point where the change occurs is zero -- this is as high as the curve gets in that local area. For a min, it is the other way around, the curve stops decreasing, turns around and starts increasing (the slope switches from negative to positive). Thus the slope at that exact point is zero.

However, you can have a place where the slope is zero, but the curve does not turn around.

So, anywhere the slope is zero is a possible (but not certain) max or min. You have to check it to make sure. You can check by graphing, or you can check using the second derivative.
• 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? • 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? • 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)? • I'm so confused.. at about Sal 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? • 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? • 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)
• Instead of calculating the double derivative for values just less than and just greater than some value, can't we find the triple derivative(the derivative of double derivative ) of that value to indicate if the concavity is going from upwards to downwards(if the triple derivative is negative) or from downwards to upwards(if the triple derivative is positive)? • We didn't actually have enough info in this video to estimate the graph. We can't be sure what happens on the ends without knowing more about X intercepts and Limits as x approaches positive and negative infinity. • We need to find:
lim as x-> negative infinity (ln(x^4+27))
lim as x-> positive infinity (ln(x^4+27))

Since x^4 is always going to be a positive number, we can simplify:
lim as |x|-> infinity (ln(x^4+27))

We can also write this as:
lim as |x|-> infinity (ln(x^4+27)) = ln (lim as |x|-> infinity (x^4+27))

As |x| approaches infinity, the 27 becomes insignificant, so:
lim as |x|-> infinity (ln(x^4+27)) = ln (lim as |x|-> infinity (x^4))

So now we are calculating the natural logarithm values closer and closer to infinity. If you calculate the natural logarithm of greater and greater input values in a calculator, you'll see that the output values will also increase.
In other words, the value of ln(x^4) will increase as the value of x^4 increases.

So:
lim as |x|-> infinity (ln(x^4+27)) = infinity
Based on this, the graph sketch in the video seems to be a good estimate of the actual graph, since the function that Sal drew appears to increase as x becomes more and more negative, or more and more positive.

I hope my explanation was clear enough.
(1 vote)
• Isn't the domain of f(x) = ln(4x + 27) be greater than 0 and not real numbers like shown here... Since, natural log of any number <= 0 is not defined....

I will greatly appreciate if someone could clear my confusion...
(1 vote) 