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Analyzing the second derivative to find inflection points

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+53x4

Step 1: Finding the second derivative
To find the inflection points of f, we need to use f:
f(x)=5x4+203x3f(x)=20x3+20x2=20x2(x+1)
Step 2: Finding all candidates
Similar to critical points, these are points where f(x)=0 or where f(x) is undefined.
f is zero at x=0 and x=1, and it's defined for all real numbers. So x=0 and x=1 are our candidates.
Step 3: Analyzing concavity
IntervalTest x-valuef(x)Conclusion
x<1x=2f(2)=80<0f is concave down
1<x<0x=0.5f(0.5)=2.5>0f is concave up
x>0x=1f(1)=40>0f is concave up
Step 4: Finding inflection points
Now that we know the intervals where f is concave up or down, we can find its inflection points (i.e. where the concavity changes direction).
  • f is concave down before x=1, concave up after it, and is defined at x=1. So f has an inflection point at x=1.
  • f is concave up before and after x=0, so it doesn't have an inflection point there.
We can verify our result by looking at the graph of f.
Function f is graphed. The x-axis goes from negative 4 to 4. The graph consists of a curve. The curve starts in quadrant 3, moves upward with decreasing steepness to about (negative 1.3, 1), moves downward with increasing steepness to about (negative 1, 0.7), continues downward with decreasing steepness to the origin, moves upward with increasing steepness, and ends in quadrant 1. The point at (negative 1, 0.7), where the graph changes from moving downward with increasing steepness to downward with decreasing steepness is the inflection point. The part of the curve to the left of this point is concave down, where the curve moves upward with decreasing steepness then downward with increasing steepness. The part of the curve to the right of the inflection point is concave up, where the curve moves downward with decreasing steepness then upward with increasing steepness.
Problem 1
Olga was asked to find where f(x)=(x2)4 has inflection points. This is her solution:
Step 1:
f(x)=4(x2)3f(x)=12(x2)2
Step 2: The solution of f(x)=0 is x=2.
Step 3: f has inflection point at x=2.
Is Olga's work correct? If not, what's her mistake?
Choose 1 answer:

Common mistake: not checking the candidates

Remember: We must not assume that any point where f(x)=0 (or where 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)=Ax3 has inflection points. This is his solution:
Step 1:
g(x)=13x23g(x)=29x53=29Ax53
Step 2: g(x)=0 has no solution.
Step 3: g doesn't have any inflection points.
Is Robert's work correct? If not, what's his mistake?
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+4x has an inflection point. This is his solution:
Step 1: h(x)=2x+4
Step 2: h(2)=0, so x=2 is a potential inflection point.
Step 3:
IntervalTest x-valueh(x)Verdict
(,2)x=3h(3)=2<0h is concave down
(2,)x=0h(0)=4>0h is concave up
Step 4: h is concave down before x=2 and concave up after x=2, so h has an inflection point at x=2.
Is Tom's work correct? If not, what's his mistake?
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+7.
For what values of x does the graph of g have a point of inflection?
Choose all answers that apply:

Want more practice? Try this exercise.

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