Answer :
To determine which of the two functions [tex]\( f(x) = -x^4 - 2 \)[/tex] and [tex]\( g(x) = -3x^3 + 2 \)[/tex] has the largest maximum [tex]\( y \)[/tex]-value, we need to find their critical points and evaluate the functions at those points.
### Finding Critical Points
First, we find the critical points by taking the derivative of each function and setting it equal to zero.
1. For [tex]\( f(x) = -x^4 - 2 \)[/tex]:
The derivative, [tex]\( f'(x) \)[/tex], is calculated as:
[tex]\[ f'(x) = \frac{d}{dx} (-x^4 - 2) = -4x^3 \][/tex]
Setting [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ -4x^3 = 0 \implies x = 0 \][/tex]
2. For [tex]\( g(x) = -3x^3 + 2 \)[/tex]:
The derivative, [tex]\( g'(x) \)[/tex], is calculated as:
[tex]\[ g'(x) = \frac{d}{dx} (-3x^3 + 2) = -9x^2 \][/tex]
Setting [tex]\( g'(x) = 0 \)[/tex]:
[tex]\[ -9x^2 = 0 \implies x = 0 \][/tex]
### Evaluating the Functions at the Critical Points
Next, evaluate each function at its critical point to determine the maximum [tex]\( y \)[/tex]-value.
1. For [tex]\( f(x) \)[/tex]:
Substitute [tex]\( x = 0 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(0) = -0^4 - 2 = -2 \][/tex]
So, the maximum [tex]\( y \)[/tex]-value for [tex]\( f(x) \)[/tex] is [tex]\( -2 \)[/tex].
2. For [tex]\( g(x) \)[/tex]:
Substitute [tex]\( x = 0 \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(0) = -3(0)^3 + 2 = 2 \][/tex]
So, the maximum [tex]\( y \)[/tex]-value for [tex]\( g(x) \)[/tex] is [tex]\( 2 \)[/tex].
### Conclusion
Comparing the maximum [tex]\( y \)[/tex]-values of both functions:
- The maximum [tex]\( y \)[/tex]-value for [tex]\( f(x) \)[/tex] is [tex]\( -2 \)[/tex].
- The maximum [tex]\( y \)[/tex]-value for [tex]\( g(x) \)[/tex] is [tex]\( 2 \)[/tex].
Hence, the function [tex]\( g(x) \)[/tex] has the largest maximum [tex]\( y \)[/tex]-value.
Therefore, the answer is:
D. [tex]\( g(x) \)[/tex]
### Finding Critical Points
First, we find the critical points by taking the derivative of each function and setting it equal to zero.
1. For [tex]\( f(x) = -x^4 - 2 \)[/tex]:
The derivative, [tex]\( f'(x) \)[/tex], is calculated as:
[tex]\[ f'(x) = \frac{d}{dx} (-x^4 - 2) = -4x^3 \][/tex]
Setting [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ -4x^3 = 0 \implies x = 0 \][/tex]
2. For [tex]\( g(x) = -3x^3 + 2 \)[/tex]:
The derivative, [tex]\( g'(x) \)[/tex], is calculated as:
[tex]\[ g'(x) = \frac{d}{dx} (-3x^3 + 2) = -9x^2 \][/tex]
Setting [tex]\( g'(x) = 0 \)[/tex]:
[tex]\[ -9x^2 = 0 \implies x = 0 \][/tex]
### Evaluating the Functions at the Critical Points
Next, evaluate each function at its critical point to determine the maximum [tex]\( y \)[/tex]-value.
1. For [tex]\( f(x) \)[/tex]:
Substitute [tex]\( x = 0 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(0) = -0^4 - 2 = -2 \][/tex]
So, the maximum [tex]\( y \)[/tex]-value for [tex]\( f(x) \)[/tex] is [tex]\( -2 \)[/tex].
2. For [tex]\( g(x) \)[/tex]:
Substitute [tex]\( x = 0 \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(0) = -3(0)^3 + 2 = 2 \][/tex]
So, the maximum [tex]\( y \)[/tex]-value for [tex]\( g(x) \)[/tex] is [tex]\( 2 \)[/tex].
### Conclusion
Comparing the maximum [tex]\( y \)[/tex]-values of both functions:
- The maximum [tex]\( y \)[/tex]-value for [tex]\( f(x) \)[/tex] is [tex]\( -2 \)[/tex].
- The maximum [tex]\( y \)[/tex]-value for [tex]\( g(x) \)[/tex] is [tex]\( 2 \)[/tex].
Hence, the function [tex]\( g(x) \)[/tex] has the largest maximum [tex]\( y \)[/tex]-value.
Therefore, the answer is:
D. [tex]\( g(x) \)[/tex]
