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AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 5
Lesson 6: Determining concavity of intervals and finding points of inflection: graphicalAnalyzing concavity (graphical)
AP.CALC:
FUN‑4 (EU)
, FUN‑4.A (LO)
, FUN‑4.A.4 (EK)
, FUN‑4.A.5 (EK)
Sal walks through an exercise where you are asked to recognize the concavity of a function in certain regions. Created by Sal Khan.
Want to join the conversation?
Is there an algebraic method for determining concavity?(7 votes)- Yes. However, it will often be less work (and less prone to error) to use the Calculus method. Anyway here is how to find concavity without calculus.
Step 1: Given f(x), find f(a), f(b), f(c), for x= a, b and c, where a < c < b
Where a and b are the points of interest. C is just any convenient point in between them.
Step 2: Find the equation of the line that connects the points found for a and b.
Step 3: Find that y-coordinate of the line from Step 2 at point C.
Step 4: If the value found in Step 3 is greater than f(c) of the original function, then you have a positive concavity. If less, you have a negative concavity. If equal, you have an average concavity of zero.
Example: f(x) = x⁵ - 5x³ + 8x² - 2 at x = 3 and 5
f(3) = 178 ; f(5) = 2698
For a point in between, I pick 4 (could be any point between 3 and 5, though). f(4) = 830
Step 2: The line connecting (3, 178), (5, 2698) is y = 1260x-3602
Step 3: At x=4, the line's y-coordinate is y = 1438
Step 4: 1438 > 830. Therefore, this function as a positive concavity between f(3) and f(5).
It should be obvious, though, that this method can make mistakes (particularly, if there is an inflection point between A and B).(32 votes)
if the 2nd derivative means concavity, then doesn't the 3rd derivative mean the frequency? (i.e. rate of change of concavity)(2 votes)
Typically in mathematics and natural sciences, we don't deal too much with the 3rd derivative. I think the easiest way to understand the 3rd derivative is through physics, in which it is the rate of change of acceleration. If you have ever driven a car and felt that your body was being sucked into the seat, then that is the feeling of acceleration, but if you can feel that you are progressively getting sucked into the seat harder or slower, then that is the jerk (3rd derivative). It's much easier to visualize in that sense, but it mathematically, the rate of change of the concavity is equivalent. Frequency typically deals with waves, so we save it for there.(19 votes)
- If there are two critical points where the slope is 0 -let's call them f(a) and f(b)- Can we always make the assumption that f''((a+b)/2) = 0? In other words, can we assume that the second derivative of a function at the average x value of two critical points always be 0? I'm just wondering because all graphs I've seen Sal use appear to have this property.(5 votes)
Interesting question – I was sure this wasn't true, but it was harder than I expected to come up with a counter example.
Try this:f(x) = sin(eˣ)
f'(x) = eˣ•cos(eˣ)
f"(x) = eˣ•[cos(eˣ) - eˣ•sin(eˣ)]f'(x) = 0:
eˣ•cos(eˣ) = 0
cos(eˣ) = 0 — since eˣ is never equal to zero
eˣ = π/2 + n•pi — where n can be any integer
x = ln(π/2 + c•pi) — where c is any non-negative integer
So, if we take the average of the first two of these and plug them into the equation above for f"(x):
a = ln(π/2)
b = ln(3π/2)
(a + b) / 2 ~= 1.00088885
f"(1.00088885) ~= -5.50761219357
I suspect that the graphs you are talking about are third order (cubic) polynomials, which thus have a constant third derivative. I believe that this constant rate of change in the second derivative leads to your observation.
Based on this, if you play around with fourth order polynomials I think you will find that your rule doesn't hold true for them either ...
EDIT:
Checked forg(x) = -x⁴ + x²– I get h"[(a+b)/2] = 0.5 ...(8 votes)
What is the the general purpose of the inflection point and the second derivative (Calculus 1)?
What are the limitations of the second derivative?
One that comes to mind, though I don't fully understand, is I tried to use f'' to to find a local max or min, but it didn't work...
Thanks(2 votes)- The second derivative is one of the higher order derivatives. One example is the second derivative of the position function, s(t), which is equivalent to taking the derivative of the velocity function, v(t). The result of doing so will yield the acceleration function, a(t). This especially useful in problems such as those concerning the gravitational forces of the earth and the moon, and finding the ratio.
To find the local maximum and minimum, I believe you must take the first derivative of the function f(x), and then set this derivative equal to zero. For example,
if f(x) = x^2-4x
then f'(x) = 2x-4
When you set this equal to zero,
2x-4 = 0
x=2
Substitution of this x value into f(x) will produce the corresponding y value, which will be -4.
Thus one can conclude that there is a minimum of -4, at x=2(3 votes)
" f''(x)>0 means the slope of f(x) is increasing "
is the meaning of that if the slope is positive it become more positive and if the slope is negative it become more negative??(2 votes)- First half is correct, but if the slope is negative and f"(x) > 0, then the slope still increases – this means it becomes less negative, i.e. becomes closer to zero.(2 votes)
I have a general question, so i learned that in an inflow/ outflow problem, when it asks for the maximum amount, you graph the rate and the constant inflow and find the intersection. Can someone explain the reason behind this?(2 votes)- If f"(x) is always greater than 0 for some interval (a,b). Does it mean that the graph of f is positive, increasing on the interval (a,b)?(1 vote)
No, f can be negative, decreasing, both, or neither on that interval. If f'' is positive, it means the graph is concave up (that is, if you fit a circle into the graph, it will be above the curve, not below it).(2 votes)
- Could this be seen as the absolute value of f'(x)'s increasing being f"(x) decreasing?(1 vote)
why does the khan academy look different in the video then actually is?(1 vote)
where do you get this practice?(1 vote)
Video transcript
A function f of x
is plotted below. Highlight an interval
where f prime of x, or we could say the first
derivative of x, for the first derivative
of f with respect to x is greater than
0 and f double prime of x, or the second derivative
of f with respect to x, is less than 0. So let's think about
what they're saying. So we're looking for a place
where the first derivative is greater than 0. That means that the slope of
the tangent line is positive. That means that the function is
increasing over that interval. So if we just think
about it here, over this whole region
right over here, the function is
clearly decreasing. Then the slope becomes
0 right over here. And then the function
starts increasing again, all the way until this
point right over here. It hits 0. And then it goes, and the
function starts decreasing. Just this first constraint
right over here tells us it's going to be something in
this interval right over there. And then they say where
the second derivative is less than 0. So this means that
the slope itself, whether it's
positive or negative, that it's actually decreasing. We are going to be concave
downwards right over here. The slope itself--
it could be positive. But it will be becoming less
and less and less positive. And so we're looking for a place
where the slope is positive, but it's becoming less and
less and less positive. If you look over here,
the slope is positive. But the slope is increasing. It's getting steeper and
steeper and steeper as we go. And then, all of a
sudden, it starts getting less steep, less
steep, less steep, less steep all the way to when the
slope gets back to 0. So if we want to
select an interval, it would be this
interval right over here. Our slope is positive. Our function is
clearly increasing, but it is increasing at
a lower and lower rate. So I will select that
right over there. Let's do one more example. A function f of x
is plotted below. Highlight an interval where f
prime of x is greater than 0. So the same thing where
our function is increasing, but it's increasing at a
slower and slower rate. So our function is increasing
in this whole region right over here, and we
see it's really steep here, that it's getting less
steep and less steep. And it's getting
closer and closer to 0, the slope of
the tangent line or the rate of increase
of the function. So I would pick anything right
around this region right here.