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

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.

why do we have to check our candidates to see if the function is defined at those points?

NOt checking to see if defined. Checking to see if the candidate is an inflection point. Review the definition of inflection point.

weren't you supposed to use chain rule in Olga's calculation to find the derivative of (x-2)^4

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?

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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

Q: why do we have to check our candidates to see if the function is defined at those points?

NOt checking to see if defined. Checking to see if the candidate is an inflection point. Review the definition of inflection point.

Google Classroom

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, 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

Interval	Test x-value	f, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis	Conclusion
x, is less than, minus, 1	x, equals, minus, 2	f, start superscript, prime, prime, end superscript, left parenthesis, minus, 2, right parenthesis, equals, minus, 80, is less than, 0	f is concave down \cap
minus, 1, is less than, x, is less than, 0	x, equals, minus, 0, point, 5	f, start superscript, prime, prime, end superscript, left parenthesis, minus, 0, point, 5, right parenthesis, equals, 2, point, 5, is greater than, 0	f is concave up \cup
x, is greater than, 0	x, equals, 1	f, start superscript, prime, prime, end superscript, left parenthesis, 1, right parenthesis, equals, 40, is greater than, 0	f 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?

(Choice A)
Olga's work is correct.
(Choice B)
Step 1 is incorrect. Olga didn't differentiate f, prime correctly.
(Choice C)
Step 2 is incorrect. x, equals, 2 isn't a solution of f, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis, equals, 0.
(Choice D)
Step 3 is incorrect. x, equals, 2 is just a candidate.

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?

(Choice A)
Robert's work is correct.
(Choice B)
Step 1 is incorrect. Robert didn't differentiate g correctly.
(Choice C)
Step 2 is incorrect. g, start superscript, prime, prime, end superscript, left parenthesis, x, right parenthesis, equals, 0 does have a solution.
(Choice D)
Step 3 is incorrect. Robert didn't consider the point where g, start superscript, prime, prime, end superscript is undefined.

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:

Interval	Test x-value	h, prime, left parenthesis, x, right parenthesis	Verdict
left parenthesis, minus, infinity, comma, minus, 2, right parenthesis	x, equals, minus, 3	h, prime, left parenthesis, minus, 3, right parenthesis, equals, minus, 2, is less than, 0	h is concave down \cap
left parenthesis, minus, 2, comma, infinity, right parenthesis	x, equals, 0	h, prime, left parenthesis, 0, right parenthesis, equals, 4, is greater than, 0	h 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?

(Choice A)
Tom's work is correct.
(Choice B)
Step 1 is incorrect. Tom didn't differentiate h correctly.
(Choice C)
Step 2 is incorrect. Tom should've considered h, start superscript, prime, prime, end superscript, not h, prime.
(Choice D)
Step 4 is incorrect. An inflection point only occurs when a function goes from being concave up to being concave down.

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?

Want more practice? Try this exercise.

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Joshua Ogunmefun
Posted 6 years ago. Direct link to Joshua Ogunmefun's post “weren't you supposed to u...”
weren't you supposed to use chain rule in Olga's calculation to find the derivative of (x-2)^4
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(0 votes)
Answer
- Tor Alm
  Posted 6 years ago. Direct link to Tor Alm's post “I think, since the deriva...”
  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?
  Comment on Tor Alm's post “I think, since the deriva...”
  (38 votes)
gupta70365
Posted 4 years ago. Direct link to gupta70365's post “If f(x) = ax^4 + bx^2, ab...”
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
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(2 votes)
Answer
- Jerry Nilsson
  Posted 4 years ago. Direct link to Jerry Nilsson's post “𝑓(𝑥) = 𝑎𝑥⁴ + 𝑏𝑥² ⇒ ...”
  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.
  Button navigates to signup page
  (5 votes)
mohamad
Posted 4 years ago. Direct link to mohamad's post “why do we have to check o...”
why do we have to check our candidates to see if the function is defined at those points?
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(1 vote)
Answer
- Richard Castillo
  Posted 4 years ago. Direct link to Richard Castillo's post “NOt checking to see if de...”
  NOt checking to see if defined. Checking to see if the candidate is an inflection point. Review the definition of inflection point.
  Button navigates to signup page
  (4 votes)
williamkrohg15
Posted 6 months ago. Direct link to williamkrohg15's post “Could you go over finding...”
Could you go over finding the candidate?
Button navigates to signup pageComment on williamkrohg15's post “Could you go over finding...”
(1 vote)
Answer
kathlyngumbo5
Posted 2 years ago. Direct link to kathlyngumbo5's post “How do we find the inflec...”
How do we find the inflection points when we have a 5 polinomial function
Button navigates to signup pageComment on kathlyngumbo5's post “How do we find the inflec...”
(1 vote)
Answer
- Alexandre Rodrigues
  Posted 2 years ago. Direct link to Alexandre Rodrigues's post “You can multiply it all a...”
  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.
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  (1 vote)