Thank you for visiting In Problems 39 62 a Find the local extrema of tex f tex b Determine the intervals on which tex f tex is concave up. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Let's solve problem 42, where the function is given as [tex]\( f(x) = x^4 + 4x \)[/tex]. We will find the local extrema, intervals of concavity, and points of inflection by following these steps:
### a. Find the local extrema of [tex]\( f(x) \)[/tex].
1. First Derivative:
Find the first derivative [tex]\( f'(x) \)[/tex] to locate critical points (where the derivative is zero or undefined).
[tex]\[
f'(x) = \frac{d}{dx}(x^4 + 4x) = 4x^3 + 4
\][/tex]
2. Set the first derivative to zero:
Solve [tex]\( f'(x) = 0 \)[/tex] to find critical points.
[tex]\[
4x^3 + 4 = 0
\][/tex]
[tex]\[
x^3 = -1
\][/tex]
[tex]\[
x = -1
\][/tex]
The critical point is [tex]\( x = -1 \)[/tex].
3. Second Derivative Test:
Find the second derivative [tex]\( f''(x) \)[/tex] to determine the concavity at the critical point.
[tex]\[
f''(x) = \frac{d}{dx}(4x^3 + 4) = 12x^2
\][/tex]
Evaluate the second derivative at [tex]\( x = -1 \)[/tex].
[tex]\[
f''(-1) = 12(-1)^2 = 12 > 0
\][/tex]
Since [tex]\( f''(-1) > 0 \)[/tex], the function has a local minimum at [tex]\( x = -1 \)[/tex].
### b. Determine the intervals on which [tex]\( f(x) \)[/tex] is concave up and concave down.
1. Analyze the sign of the second derivative:
[tex]\[
f''(x) = 12x^2
\][/tex]
Since [tex]\( 12x^2 \)[/tex] is always non-negative, [tex]\( f''(x) > 0 \)[/tex] for all [tex]\( x \neq 0 \)[/tex] and [tex]\( f''(x) = 0 \)[/tex] at [tex]\( x = 0 \)[/tex].
2. Concavity:
- The function is concave up on [tex]\( (-\infty, \infty) \)[/tex] except at [tex]\( x = 0 \)[/tex] where it changes concavity (check next for inflection).
### c. Find any points of inflection.
1. Points of Inflection:
A point of inflection occurs where the second derivative changes sign. We found that [tex]\( f''(x) = 12x^2 \)[/tex].
- As [tex]\( x \)[/tex] moves through 0, there is only a change from zero to positive concavity (but no negative), suggesting a potential inflection point, but since there is no sign change around the point [tex]\( x = 0 \)[/tex] (as it moves only within the positive region by a gap), technically it doesn't change concavity.
Based on these analyses:
- Local Extrema: There is a local minimum at [tex]\( x = -1 \)[/tex].
- Intervals of Concavity: [tex]\( f(x) \)[/tex] is concave up for [tex]\( (-\infty, \infty) \)[/tex].
- Points of Inflection: Technically, [tex]\( x = 0 \)[/tex] might be a consideration point, but since there's no true sign change, it is not a typical inflection.
### a. Find the local extrema of [tex]\( f(x) \)[/tex].
1. First Derivative:
Find the first derivative [tex]\( f'(x) \)[/tex] to locate critical points (where the derivative is zero or undefined).
[tex]\[
f'(x) = \frac{d}{dx}(x^4 + 4x) = 4x^3 + 4
\][/tex]
2. Set the first derivative to zero:
Solve [tex]\( f'(x) = 0 \)[/tex] to find critical points.
[tex]\[
4x^3 + 4 = 0
\][/tex]
[tex]\[
x^3 = -1
\][/tex]
[tex]\[
x = -1
\][/tex]
The critical point is [tex]\( x = -1 \)[/tex].
3. Second Derivative Test:
Find the second derivative [tex]\( f''(x) \)[/tex] to determine the concavity at the critical point.
[tex]\[
f''(x) = \frac{d}{dx}(4x^3 + 4) = 12x^2
\][/tex]
Evaluate the second derivative at [tex]\( x = -1 \)[/tex].
[tex]\[
f''(-1) = 12(-1)^2 = 12 > 0
\][/tex]
Since [tex]\( f''(-1) > 0 \)[/tex], the function has a local minimum at [tex]\( x = -1 \)[/tex].
### b. Determine the intervals on which [tex]\( f(x) \)[/tex] is concave up and concave down.
1. Analyze the sign of the second derivative:
[tex]\[
f''(x) = 12x^2
\][/tex]
Since [tex]\( 12x^2 \)[/tex] is always non-negative, [tex]\( f''(x) > 0 \)[/tex] for all [tex]\( x \neq 0 \)[/tex] and [tex]\( f''(x) = 0 \)[/tex] at [tex]\( x = 0 \)[/tex].
2. Concavity:
- The function is concave up on [tex]\( (-\infty, \infty) \)[/tex] except at [tex]\( x = 0 \)[/tex] where it changes concavity (check next for inflection).
### c. Find any points of inflection.
1. Points of Inflection:
A point of inflection occurs where the second derivative changes sign. We found that [tex]\( f''(x) = 12x^2 \)[/tex].
- As [tex]\( x \)[/tex] moves through 0, there is only a change from zero to positive concavity (but no negative), suggesting a potential inflection point, but since there is no sign change around the point [tex]\( x = 0 \)[/tex] (as it moves only within the positive region by a gap), technically it doesn't change concavity.
Based on these analyses:
- Local Extrema: There is a local minimum at [tex]\( x = -1 \)[/tex].
- Intervals of Concavity: [tex]\( f(x) \)[/tex] is concave up for [tex]\( (-\infty, \infty) \)[/tex].
- Points of Inflection: Technically, [tex]\( x = 0 \)[/tex] might be a consideration point, but since there's no true sign change, it is not a typical inflection.
Thank you for reading the article In Problems 39 62 a Find the local extrema of tex f tex b Determine the intervals on which tex f tex is concave up. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
- You are operating a recreational vessel less than 39 4 feet long on federally controlled waters Which of the following is a legal sound device
- Which step should a food worker complete to prevent cross contact when preparing and serving an allergen free meal A Clean and sanitize all surfaces
- For one month Siera calculated her hometown s average high temperature in degrees Fahrenheit She wants to convert that temperature from degrees Fahrenheit to degrees
Rewritten by : Jeany
