To solve the problem, let's break it down step by step.
### a) Finding the Critical Values
1. Start with the original function:
[tex]\[
g(x) = -x^3 + 12x - 7
\][/tex]
2. Find the first derivative [tex]\( g'(x) \)[/tex]:
[tex]\[
g'(x) = \frac{d}{dx} (-x^3 + 12x - 7) = -3x^2 + 12
\][/tex]
3. Set the first derivative equal to zero to find critical values:
[tex]\[
-3x^2 + 12 = 0
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
[tex]\[
-3x^2 + 12 = 0 \implies -3x^2 = -12 \implies x^2 = 4 \implies x = \pm 2
\][/tex]
So, the critical values are [tex]\( x = -2 \)[/tex] and [tex]\( x = 2 \)[/tex].
### b) Determining the Nature of Each Critical Point (Relative Extrema)
1. Find the second derivative [tex]\( g''(x) \)[/tex]:
[tex]\[
g''(x) = \frac{d}{dx} (-3x^2 + 12) = -6x
\][/tex]
2. Analyze the second derivative at the critical values:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[
g''(-2) = -6(-2) = 12 \quad (\text{positive, indicating a relative minimum})
\][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[
g''(2) = -6(2) = -12 \quad (\text{negative, indicating a relative maximum})
\][/tex]
### Summary
- Critical values:
[tex]\[
x = -2, 2
\][/tex]
- Relative extrema:
- [tex]\( x = -2 \)[/tex] is a relative minimum.
- [tex]\( x = 2 \)[/tex] is a relative maximum.
### Answer:
a) The critical values are [tex]\( -2 \)[/tex] and [tex]\( 2 \)[/tex].
[tex]\[
A. The critical value(s) of the function is/are \boxed{-2, 2}.
\][/tex]
b) The relative extrema are:
- [tex]\( x = -2 \)[/tex] is a relative minimum.
- [tex]\( x = 2 \)[/tex] is a relative maximum.
