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.
