If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Calculus AB

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

Lesson 7: Determining concavity of intervals and finding points of inflection: algebraic

# 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 $f(x)=x^5+\dfrac53x^4$f, left parenthesis, x, right parenthesis, equals, x, start superscript, 5, end superscript, plus, start fraction, 5, divided by, 3, end fraction, x, start superscript, 4, end superscript

Step 1: Finding the second derivative
To find the inflection points of f, we need to use f, start superscript, prime, prime, end superscript:
\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, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis, equals, 0 or where f, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis is undefined.
f, start superscript, prime, prime, end superscript is zero at x, equals, 0 and x, equals, minus, 1, and it's defined for all real numbers. So x, equals, 0 and x, equals, minus, 1 are our candidates.
Step 3: Analyzing concavity
IntervalTest x-valuef, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesisConclusion
x, is less than, minus, 1x, equals, minus, 2f, start superscript, prime, prime, end superscript, left parenthesis, minus, 2, right parenthesis, equals, minus, 80, is less than, 0f is concave down \cap
minus, 1, is less than, x, is less than, 0x, equals, minus, 0, point, 5f, start superscript, prime, prime, end superscript, left parenthesis, minus, 0, point, 5, right parenthesis, equals, 2, point, 5, is greater than, 0f is concave up \cup
x, is greater than, 0x, equals, 1f, start superscript, prime, prime, end superscript, left parenthesis, 1, right parenthesis, equals, 40, is greater than, 0f is concave up \cup
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, equals, minus, 1, concave up after it, and is defined at x, equals, minus, 1. So f has an inflection point at x, equals, minus, 1.
• f is concave up before and after x, equals, 0, so it doesn't have an inflection point there.
We can verify our result by looking at the graph of f.
Problem 1
Olga was asked to find where f, left parenthesis, x, right parenthesis, equals, left parenthesis, x, minus, 2, right parenthesis, start superscript, 4, end superscript has inflection points. This is her solution:
Step 1:
\begin{aligned} f'(x)&=4(x-2)^3 \\\\\\ f''(x)&=12(x-2)^2 \end{aligned}
Step 2: The solution of f, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis, equals, 0 is x, equals, 2.
Step 3: f has inflection point at x, equals, 2.
Is Olga's work correct? If not, what's her mistake?

### Common mistake: not checking the candidates

Remember: We must not assume that any point where f, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis, equals, 0 (or where f, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis 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, left parenthesis, x, right parenthesis, equals, cube root of, x, end cube root has inflection points. This is his solution:
Step 1:
\begin{aligned} g'(x)&=\dfrac13x^{-\frac23} \\\\\\ g''(x)&=-\dfrac29x^{-\frac53} \\\\ &=-\dfrac{2}{9\sqrt{x^5}} \end{aligned}
Step 2: g, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis, equals, 0 has no solution.
Step 3: g doesn't have any inflection points.
Is Robert's work correct? If not, what's his mistake?

### 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, left parenthesis, x, right parenthesis, equals, x, squared, plus, 4, x has an inflection point. This is his solution:
Step 1: h, prime, left parenthesis, x, right parenthesis, equals, 2, x, plus, 4
Step 2: h, prime, left parenthesis, minus, 2, right parenthesis, equals, 0, so x, equals, minus, 2 is a potential inflection point.
Step 3:
IntervalTest x-valueh, prime, left parenthesis, x, right parenthesisVerdict
left parenthesis, minus, infinity, comma, minus, 2, right parenthesisx, equals, minus, 3h, prime, left parenthesis, minus, 3, right parenthesis, equals, minus, 2, is less than, 0h is concave down \cap
left parenthesis, minus, 2, comma, infinity, right parenthesisx, equals, 0h, prime, left parenthesis, 0, right parenthesis, equals, 4, is greater than, 0h is concave up \cup
Step 4: h is concave down before x, equals, minus, 2 and concave up after x, equals, minus, 2, so h has an inflection point at x, equals, minus, 2.
Is Tom's work correct? If not, what's his mistake?