Let's analyze the functions [tex]\( f(x) = 4(x-6)^4 + 1 \)[/tex] and [tex]\( g(x) = 2x^3 + 28 \)[/tex] to determine which has the smallest minimum [tex]\( y \)[/tex]-value.
### Step-by-Step Solution:
#### Step 1: Identify Critical Points
##### For [tex]\( f(x) = 4(x-6)^4 + 1 \)[/tex]:
1. Find the first derivative of [tex]\( f(x) \)[/tex]:
[tex]\[
f'(x) = \frac{d}{dx}[4(x-6)^4 + 1] = 16(x-6)^3
\][/tex]
2. Set the first derivative equal to zero to find the critical points:
[tex]\[
16(x-6)^3 = 0 \quad \Rightarrow \quad x-6 = 0 \quad \Rightarrow \quad x = 6
\][/tex]
##### For [tex]\( g(x) = 2x^3 + 28 \)[/tex]:
1. Find the first derivative of [tex]\( g(x) \)[/tex]:
[tex]\[
g'(x) = \frac{d}{dx}[2x^3 + 28] = 6x^2
\][/tex]
2. Set the first derivative equal to zero to find the critical points:
[tex]\[
6x^2 = 0 \quad \Rightarrow \quad x^2 = 0 \quad \Rightarrow \quad x = 0
\][/tex]
#### Step 2: Determine the Nature of the Critical Points
##### For [tex]\( f(x) = 4(x-6)^4 + 1 \)[/tex]:
1. Find the second derivative of [tex]\( f(x) \)[/tex]:
[tex]\[
f''(x) = \frac{d}{dx}[16(x-6)^3] = 48(x-6)^2
\][/tex]
2. Evaluate the second derivative at the critical point [tex]\( x = 6 \)[/tex]:
[tex]\[
f''(6) = 48(6-6)^2 = 48 \cdot 0^2 = 0
\][/tex]
Since the second derivative is zero, the test is inconclusive. However, given the form of the function [tex]\( (x-6)^4 \)[/tex], which is always non-negative, the derivative test suggests this might be an inflection point rather than a minimum or maximum.
##### For [tex]\( g(x) = 2x^3 + 28 \)[/tex]:
1. Find the second derivative of [tex]\( g(x) \)[/tex]:
[tex]\[
g''(x) = \frac{d}{dx}[6x^2] = 12x
\][/tex]
2. Evaluate the second derivative at the critical point [tex]\( x = 0 \)[/tex]:
[tex]\[
g''(0) = 12 \cdot 0 = 0
\][/tex]
As with [tex]\( f(x) \)[/tex], the second derivative is zero, making the test inconclusive.
#### Step 3: Evaluate the Function at Critical Points
For [tex]\( f(x) = 4(x-6)^4 + 1 \)[/tex] at [tex]\( x = 6 \)[/tex]:
1. Evaluate [tex]\( f(6) \)[/tex]:
[tex]\[
f(6) = 4(6-6)^4 + 1 = 4 \cdot 0 + 1 = 1
\][/tex]
For [tex]\( g(x) = 2x^3 + 28 \)[/tex] at [tex]\( x = 0 \)[/tex]:
1. Evaluate [tex]\( g(0) \)[/tex]:
[tex]\[
g(0) = 2 \cdot 0^3 + 28 = 0 + 28 = 28
\][/tex]
#### Step 4: Compare Minimum Values
The minimum value of [tex]\( f(x) \)[/tex] is 1 at [tex]\( x = 6 \)[/tex].
The minimum value of [tex]\( g(x) \)[/tex] is 28 at [tex]\( x = 0 \)[/tex].
### Conclusion
Since [tex]\( f(x) \)[/tex] achieves a smaller minimum value (1) compared to [tex]\( g(x) \)[/tex] (28), it seems like [tex]\( f(x) \)[/tex] has the smallest minimum [tex]\( y \)[/tex]-value. However, since the second derivatives at critical points for both functions are zero, it introduces some uncertainty in inferring the correct nature of these critical points. Hence, it appears there might not be enough information to conclusively determine this without further analysis.
Thus, the correct choice is:
C. There is not enough information to determine
