Review your knowledge of inflection points and how we use differential calculus to find them.

What are inflection points?

Inflection points (or points of inflection) are points where the graph of a function changes concavity (from \cup to \cap or vice versa).
Want to learn more about inflection points and differential calculus? Check out this video.

Practice set 1: Analyzing inflection points graphically

Problem 1.1
How many inflection points does the graph of ff have?
Choose 1 answer:
Choose 1 answer:

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Practice set 2: Analyzing inflection points algebraically

Inflection points are found in a way similar to how we find extremum points. However, instead of looking for points where the derivative changes its sign, we are looking for points where the second derivative changes its sign.
Let's find, for example, the inflection points of f(x)=12x4+x36x2f(x)=\dfrac{1}{2}x^4+x^3-6x^2.
The second derivative of ff is f(x)=6(x1)(x+2)f''(x)=6(x-1)(x+2).
First, we differentiate ff to find ff':
=f(x)=ddx(12x4+x36x2)=12ddx(x4)+ddx(x3)6ddx(x2)=12(4x3)+(3x2)6(2x)=2x3+3x212x\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}(\dfrac{1}{2}x^4+x^3-6x^2) \\\\ &=\dfrac{1}{2}\dfrac{d}{dx}(x^4)+\dfrac{d}{dx}(x^3)-6\dfrac{d}{dx}(x^2) \\\\ &=\dfrac{1}{2}(4x^3)+(3x^2)-6(2x) \\\\ &=2x^3+3x^2-12x \end{aligned}
Now we can differentiate ff' to find ff'':
=f(x)=ddx(2x3+3x212x)=2ddx(x3)+3ddx(x2)12ddx(x)=2(3x2)+3(2x)12(1)=6x2+6x12=6(x2+x2)=6(x+2)(x1)\begin{aligned} &\phantom{=}f''(x) \\\\ &=\dfrac{d}{dx}(2x^3+3x^2-12x) \\\\ &=2\dfrac{d}{dx}(x^3)+3\dfrac{d}{dx}(x^2)-12\dfrac{d}{dx}(x) \\\\ &=2(3x^2)+3(2x)-12(1) \\\\ &=6x^2+6x-12 \\\\ &=6(x^2+x-2) \\\\ &=6(x+2)(x-1) \end{aligned}
f(x)=0f''(x)=0 for x=2,1x=-2,1, and it's defined everywhere. x=2x=-2 and x=1x=1 divide the number line into three intervals:
Let's evaluate ff'' at each interval to see if it's positive or negative on that interval.
Intervalxx-valuef(x)f''(x)Verdict
x<2x<-2x=3x=-3f(3)=24>0f''(-3)=24>0ff is concave up \cup
2<x<1-2<x<1x=0x=0f(0)=12<0f''(0)=-12<0ff is concave down \cap
x>1x>1x=2x=2f(2)=24>0f''(2)=24>0ff is concave up \cup
We can see that the graph of ff changes concavity at both x=2x=-2 and x=1x=1, so ff has inflection points at both of those xx-values.
Problem 2.1
g(x)=x4+4x318x2g(x)=x^4+4x^3-18x^2
For what values of xx does the graph of gg have a point of inflection?
Choose all answers that apply:
Choose all answers that apply:

Want to try more problems like this? Check out this exercise.
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