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Let's determine which of the two functions [tex]\( f(x) = x^4 - 2 \)[/tex] and [tex]\( g(x) = 3x^3 + 2 \)[/tex] has the smallest minimum [tex]\( y \)[/tex]-value by analyzing their critical points. Here is a detailed step-by-step solution:
### Step 1: Find the critical points of [tex]\( f(x) \)[/tex]
1. Compute the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^4 - 2) = 4x^3 \][/tex]
2. Set the derivative equal to zero to find the critical points:
[tex]\[ 4x^3 = 0 \implies x = 0 \][/tex]
So, the critical point of [tex]\( f(x) \)[/tex] is [tex]\( x = 0 \)[/tex].
### Step 2: Evaluate [tex]\( f(x) \)[/tex] at its critical point
1. Substitute [tex]\( x = 0 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(0) = 0^4 - 2 = -2 \][/tex]
Thus, the minimum [tex]\( y \)[/tex]-value of [tex]\( f(x) \)[/tex] is [tex]\( -2 \)[/tex].
### Step 3: Find the critical points of [tex]\( g(x) \)[/tex]
1. Compute the derivative of [tex]\( g(x) \)[/tex]:
[tex]\[ g'(x) = \frac{d}{dx}(3x^3 + 2) = 9x^2 \][/tex]
2. Set the derivative equal to zero to find the critical points:
[tex]\[ 9x^2 = 0 \implies x = 0 \][/tex]
So, the critical point of [tex]\( g(x) \)[/tex] is [tex]\( x = 0 \)[/tex].
### Step 4: Evaluate [tex]\( g(x) \)[/tex] at its critical point
1. Substitute [tex]\( x = 0 \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(0) = 3 \cdot 0^3 + 2 = 2 \][/tex]
Thus, the minimum [tex]\( y \)[/tex]-value of [tex]\( g(x) \)[/tex] is [tex]\( 2 \)[/tex].
### Step 5: Compare the minimum [tex]\( y \)[/tex]-values
The minimum [tex]\( y \)[/tex]-value of [tex]\( f(x) \)[/tex] is [tex]\( -2 \)[/tex], and the minimum [tex]\( y \)[/tex]-value of [tex]\( g(x) \)[/tex] is [tex]\( 2 \)[/tex]. Since [tex]\( -2 < 2 \)[/tex], the function [tex]\( f(x) \)[/tex] has the smallest minimum [tex]\( y \)[/tex]-value.
### Conclusion
The correct answer is:
B. [tex]\( f(x) \)[/tex]
### Step 1: Find the critical points of [tex]\( f(x) \)[/tex]
1. Compute the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^4 - 2) = 4x^3 \][/tex]
2. Set the derivative equal to zero to find the critical points:
[tex]\[ 4x^3 = 0 \implies x = 0 \][/tex]
So, the critical point of [tex]\( f(x) \)[/tex] is [tex]\( x = 0 \)[/tex].
### Step 2: Evaluate [tex]\( f(x) \)[/tex] at its critical point
1. Substitute [tex]\( x = 0 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(0) = 0^4 - 2 = -2 \][/tex]
Thus, the minimum [tex]\( y \)[/tex]-value of [tex]\( f(x) \)[/tex] is [tex]\( -2 \)[/tex].
### Step 3: Find the critical points of [tex]\( g(x) \)[/tex]
1. Compute the derivative of [tex]\( g(x) \)[/tex]:
[tex]\[ g'(x) = \frac{d}{dx}(3x^3 + 2) = 9x^2 \][/tex]
2. Set the derivative equal to zero to find the critical points:
[tex]\[ 9x^2 = 0 \implies x = 0 \][/tex]
So, the critical point of [tex]\( g(x) \)[/tex] is [tex]\( x = 0 \)[/tex].
### Step 4: Evaluate [tex]\( g(x) \)[/tex] at its critical point
1. Substitute [tex]\( x = 0 \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(0) = 3 \cdot 0^3 + 2 = 2 \][/tex]
Thus, the minimum [tex]\( y \)[/tex]-value of [tex]\( g(x) \)[/tex] is [tex]\( 2 \)[/tex].
### Step 5: Compare the minimum [tex]\( y \)[/tex]-values
The minimum [tex]\( y \)[/tex]-value of [tex]\( f(x) \)[/tex] is [tex]\( -2 \)[/tex], and the minimum [tex]\( y \)[/tex]-value of [tex]\( g(x) \)[/tex] is [tex]\( 2 \)[/tex]. Since [tex]\( -2 < 2 \)[/tex], the function [tex]\( f(x) \)[/tex] has the smallest minimum [tex]\( y \)[/tex]-value.
### Conclusion
The correct answer is:
B. [tex]\( f(x) \)[/tex]
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