Learn how the second derivative of a function is used in order to find the function's inflection points. Learn which common mistakes to avoid in the process.
We can find the inflection points of a function by analyzing its second derivative.

Example: Finding the inflection points of f(x)=x5+53x4f(x)=x^5+\dfrac53x^4

Step 1: Finding the second derivative
To find the inflection points of ff, we need to use ff'':
f(x)=5x4+203x3f(x)=20x3+20x2=20x2(x+1)\begin{aligned} f'(x)&=5x^4+\dfrac{20}{3}x^3 \\\\ f''(x)&=20x^3+20x^2 \\\\ &=20x^2(x+1) \end{aligned}
Step 2: Finding all candidates
Similar to critical points, these are points where f(x)=0f''(x)=0 or where f(x)f''(x) is undefined.
ff'' is zero at x=0x=0 and x=1x=-1, and it's defined for all real numbers. So x=0x=0 and x=1x=-1 are our candidates.
Step 3: Analyzing concavity
IntervalTest xx-valuef(x)f''(x)Conclusion
x<1x<-1x=2x=-2f(2)=80<0f''(-2)=-80<0ff is concave down \cap
1<x<0-1<x<0x=0.5x=-0.5f(0.5)=2.5>0f''(-0.5)=2.5>0ff is concave up \cup
x>0x>0x=1x=1f(1)=40>0f''(1)=40>0ff is concave up \cup
Step 4: Finding inflection points
Now that we know the intervals where ff is concave up or down, we can find its inflection points (i.e. where the concavity changes direction).
  • ff is concave down before x=1x=-1, concave up after it, and is defined at x=1x=-1. So ff has an inflection point at x=1x=-1.
  • ff is concave up before and after x=0x=0, so it doesn't have an inflection point there.
We can verify our result by looking at the graph of ff.
Problem 1
Olga was asked to find where f(x)=(x2)4f(x)=(x-2)^4 has inflection points. This is her solution:
Step 1:
f(x)=4(x2)3f(x)=12(x2)2\begin{aligned} f'(x)&=4(x-2)^3 \\\\\\ f''(x)&=12(x-2)^2 \end{aligned}
Step 2: The solution of f(x)=0f''(x)=0 is x=2x=2.
Step 3: ff has inflection point at x=2x=2.
Is Olga's work correct? If not, what's her mistake?
Choose 1 answer:
Choose 1 answer:

Common mistake: not checking the candidates

Remember: We must not assume that any point where f(x)=0f''(x)=0 (or where f(x)f''(x) is undefined) is an inflection point. Instead, we should check our candidates to see if the second derivative changes signs at those points and the function is defined at those points.
Problem 2
Robert was asked to find where g(x)=x3g(x)=\sqrt[3]{x} has inflection points. This is his solution:
Step 1:
g(x)=13x23g(x)=29x53=29x53\begin{aligned} g'(x)&=\dfrac13x^{-\frac23} \\\\\\ g''(x)&=-\dfrac29x^{-\frac53} \\\\ &=-\dfrac{2}{9\sqrt[3]{x^5}} \end{aligned}
Step 2: g(x)=0g''(x)=0 has no solution.
Step 3: gg doesn't have any inflection points.
Is Robert's work correct? If not, what's his mistake?
Choose 1 answer:
Choose 1 answer:

Common mistake: not including points where the derivative is undefined

Remember: Our candidates for inflection points are points where the second derivative is equal to zero and points where the second derivative is undefined. Ignoring points where the second derivative is undefined will often result in a wrong answer.
Problem 3
Tom was asked to find whether h(x)=x2+4xh(x)=x^2+4x has an inflection point. This is his solution:
Step 1: h(x)=2x+4h'(x)=2x+4
Step 2: h(2)=0h'(-2)=0, so x=2x=-2 is a potential inflection point.
Step 3:
IntervalTest xx-valueh(x)h'(x)Verdict
(,2)(-\infty,-2)x=3x=-3h(3)=2<0h'(-3)=-2<0hh is concave down \cap
(2,)(-2,\infty)x=0x=0h(0)=4>0h'(0)=4>0hh is concave up \cup
Step 4: hh is concave down before x=2x=-2 and concave up after x=2x=-2, so hh has an inflection point at x=2x=-2.
Is Tom's work correct? If not, what's his mistake?
Choose 1 answer:
Choose 1 answer:

Common mistake: looking at the first derivative instead of the second derivative

Remember: When looking for inflection points, we must always analyze where the second derivative changes its sign. Doing this for the first derivative will give us relative extremum points, not inflection points.
Problem 4
Let g(x)=x412x342x2+7g(x)=x^4-12x^3-42x^2+7.
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 more practice? Try this exercise.
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