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Which of the two functions below has the smallest minimum [tex]$y$[/tex]-value?

[tex]\[
\begin{array}{c}
f(x) = x^4 - 2 \\
g(x) = 3x^3 + 2
\end{array}
\][/tex]

A. [tex]\( f(x) \)[/tex]
B. There is not enough information to determine
C. [tex]\( g(x) \)[/tex]
D. The extreme minimum [tex]\( y \)[/tex]-value for [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] is [tex]\(-\infty\)[/tex]



Answer :

GinnyAnswer
To determine which function has the smallest minimum [tex]\( y \)[/tex]-value, we need to follow these steps:

1. Find the first derivative of each function: The first derivative helps us locate the critical points, which are candidates for minimum or maximum values.

[tex]\[ f(x) = x^4 - 2 \][/tex]
[tex]\[ f'(x) = \frac{d}{dx}(x^4 - 2) = 4x^3 \][/tex]

[tex]\[ g(x) = 3x^3 + 2 \][/tex]
[tex]\[ g'(x) = \frac{d}{dx}(3x^3 + 2) = 9x^2 \][/tex]

2. Set the first derivatives to zero to find the critical points:

[tex]\[ f'(x) = 4x^3 = 0 \][/tex]
[tex]\[ x = 0 \][/tex]

[tex]\[ g'(x) = 9x^2 = 0 \][/tex]
[tex]\[ x = 0 \][/tex]

3. Evaluate the original functions at the critical points to determine the [tex]\( y \)[/tex]-values:

For [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:

[tex]\[ f(0) = 0^4 - 2 = -2 \][/tex]

For [tex]\( g(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:

[tex]\[ g(0) = 3(0^3) + 2 = 2 \][/tex]

4. Compare the minimum [tex]\( y \)[/tex]-values:

- The minimum [tex]\( y \)[/tex]-value for [tex]\( f(x) \)[/tex] is [tex]\(-2\)[/tex].
- The minimum [tex]\( y \)[/tex]-value for [tex]\( g(x) \)[/tex] is [tex]\(2\)[/tex].

Thus, the function [tex]\( f(x) = x^4 - 2 \)[/tex] has the smallest minimum [tex]\( y \)[/tex]-value among the two given functions.

The correct answer is:
A. [tex]\( f(x) \)[/tex]

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