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## AP®︎/College Calculus BC

### Unit 5: Lesson 7

Determining concavity of intervals and finding points of inflection: algebraic

# Analyzing the second derivative to find inflection points

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.A (LO)
,
FUN‑4.A.4 (EK)
,
FUN‑4.A.5 (EK)
,
FUN‑4.A.6 (EK)
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.
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, 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?
Choose 1 answer:
Choose 1 answer:

### 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[3]{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?
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, 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?
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, left parenthesis, x, right parenthesis, equals, x, start superscript, 4, end superscript, minus, 12, x, cubed, minus, 42, x, squared, plus, 7.
For what values of x does the graph of g have a point of inflection?
Choose all answers that apply:
Choose all answers that apply:

Want more practice? Try this exercise.

## Want to join the conversation?

• weren't you supposed to use chain rule in Olga's calculation to find the derivative of (x-2)^4
(0 votes)
• I think, since the derivative of (X-2) is simply just 1 all the time, they don't write it out.

Example:

g(x) = (x-2)^4

g'(x)=4 * (x-2)^3 * 1 <--- Since the derivative of (x-2) is just 1, you don't have to write it out. Did i understand your question correctly?
(36 votes)
• If f(x) = ax^4 + bx^2, ab>0, then which of the following is correct:
1. The curve has no horizontal tangents
2. The curve is concave up for all x
3. The curve is concave down for all x
4. The curve has no inflection point
5. None of the preceding is necessarily true
(2 votes)
• 𝑓(𝑥) = 𝑎𝑥⁴ + 𝑏𝑥² ⇒
⇒ 𝑓 '(𝑥) = 4𝑎𝑥³ + 2𝑏𝑥 ⇒
⇒ 𝑓 ''(𝑥) = 12𝑎𝑥² + 2𝑏

𝑓 ''(𝑥) = 0 ⇒ 𝑥² = −𝑏∕(6𝑎), which doesn't have any real number solutions since 𝑎𝑏 > 0 (𝑎 and 𝑏 are either both positive or both negative), and so 𝑓(𝑥) has no inflection points.

Thereby, 𝑓(𝑥) is either always concave up or always concave down, which means that it can only have one local extreme point, and that point must be (0, 0) because 𝑥 = 0 obviously solves 𝑓 '(𝑥) = 0 (which by the way tells us that 𝑓(𝑥) does have a horizontal tangent).

Since the local extremum is 𝑓(0) = 0, all we need to do in order to find the concavity is to check whether 𝑓(𝑥) is positive or negative for any 𝑥 ≠ 0,
for example 𝑥 = 1 ⇒ 𝑓(1) = 𝑎 + 𝑏, which would be positive for 𝑎, 𝑏 > 0 and negative for 𝑎, 𝑏 < 0.
So, without knowing the sign of 𝑎 and 𝑏 we can't tell whether 𝑓(𝑥) is concave up or down.
(4 votes)
• why do we have to check our candidates to see if the function is defined at those points?
(1 vote)
• NOt checking to see if defined. Checking to see if the candidate is an inflection point. Review the definition of inflection point.
(4 votes)
• Could you go over finding the candidate?
(1 vote)
• How do we find the inflection points when we have a 5 polinomial function
(1 vote)
• You can multiply it all and use the power rule like charlie said, you can also use a mix of the power rule and the chain rule when differentiating if the function is in parentheses. It doesn't matter what the intial polynomial is, to find the inflection points you always need to use the second derivative.
(1 vote)