Certainly! Let's solve the given problem step by step.
Given the function:
[tex]\[ g(x) = -x^3 + 3x - 10 \][/tex]
### Step 1: Finding the Critical Values
Critical values occur where the first derivative of the function is zero or where it is undefined.
#### Calculate the first derivative [tex]\( g'(x) \)[/tex]:
[tex]\[ g'(x) = \frac{d}{dx}(-x^3 + 3x - 10) \][/tex]
[tex]\[ g'(x) = -3x^2 + 3 \][/tex]
#### Set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ -3x^2 + 3 = 0 \][/tex]
[tex]\[ -3(x^2 - 1) = 0 \][/tex]
[tex]\[ -3(x-1)(x+1) = 0 \][/tex]
This gives us the critical points:
[tex]\[ x - 1 = 0 \quad \text{or} \quad x + 1 = 0 \][/tex]
[tex]\[ x = 1 \quad \text{or} \quad x = -1 \][/tex]
So, the critical values are:
[tex]\[ x = 1, -1 \][/tex]
Thus, the answer to part (a) is:
A. The critical values of the function are [tex]\( 1, -1 \)[/tex].
### Step 2: Finding the Relative Extrema
We need to determine the nature of each critical point (whether it's a relative maximum, minimum, or neither). To do this, we examine the second derivative test.
#### Calculate the second derivative [tex]\( g''(x) \)[/tex]:
[tex]\[ g''(x) = \frac{d}{dx}(-3x^2 + 3) \][/tex]
[tex]\[ g''(x) = -6x \][/tex]
Evaluate the second derivative at each critical point:
At [tex]\( x = 1 \)[/tex]:
[tex]\[ g''(1) = -6(1) = -6 \][/tex]
Since [tex]\( g''(1) < 0 \)[/tex], the function has a relative maximum at [tex]\( x = 1 \)[/tex].
At [tex]\( x = -1 \)[/tex]:
[tex]\[ g''(-1) = -6(-1) = 6 \][/tex]
Since [tex]\( g''(-1) > 0 \)[/tex], the function has a relative minimum at [tex]\( x = -1 \)[/tex].
Thus, the relative extrema are:
- A relative maximum at [tex]\( x = 1 \)[/tex].
- A relative minimum at [tex]\( x = -1 \)[/tex].
### Summary:
a) The critical values of the function are [tex]\( 1, -1 \)[/tex].
b) The relative extrema are:
- A relative maximum at [tex]\( x = 1 \)[/tex].
- A relative minimum at [tex]\( x = -1 \)[/tex].
