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Let [tex]f(x) = (6 - 4x)^6[/tex].

1. [tex]f(x)[/tex] has one critical value at [tex]A = \square[/tex].

2. For [tex]x \ \textless \ A[/tex], [tex]f(x)[/tex] is [Select an answer].

3. For [tex]x \ \textgreater \ A[/tex], [tex]f(x)[/tex] is [Select an answer].

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GinnyAnsweridn
To solve for the critical points of the function [tex]\( f(x) = (6 - 4x)^6 \)[/tex] and determine the behavior of the function around these points, let’s go through the following steps:

### Step 1: Find the derivative of [tex]\( f(x) \)[/tex]
We begin by taking the derivative of the function.

[tex]\[ f(x) = (6 - 4x)^6 \][/tex]

We apply the chain rule to find [tex]\( f'(x) \)[/tex]:

[tex]\[ f'(x) = 6 \cdot (6 - 4x)^5 \cdot (-4) \][/tex]
[tex]\[ f'(x) = -24 \cdot (6 - 4x)^5 \][/tex]

### Step 2: Find the critical points
Critical points occur where the derivative is equal to zero. Set [tex]\( f'(x) = 0 \)[/tex]:

[tex]\[ -24 \cdot (6 - 4x)^5 = 0 \][/tex]

Since a non-zero constant multiplied by an expression cannot be zero, we focus on the term [tex]\( (6 - 4x)^5 = 0 \)[/tex]:

[tex]\[ (6 - 4x) = 0 \][/tex]
[tex]\[ 6 = 4x \][/tex]
[tex]\[ x = \frac{6}{4} \][/tex]
[tex]\[ x = 1.5 \][/tex]

Thus, the critical value is at [tex]\( A = 1.5 \)[/tex].

### Step 3: Determine the behavior of [tex]\( f(x) \)[/tex] around [tex]\( x = 1.5 \)[/tex]
To understand the behavior of the function around the critical point, we need to examine the sign of the derivative just to the left and just to the right of [tex]\( x = 1.5 \)[/tex].

- For [tex]\( x < 1.5 \)[/tex]:

Select a point slightly less than 1.5, say [tex]\( x = 1.4 \)[/tex]:

[tex]\[ f'(x) = -24 \cdot (6 - 4 \cdot 1.4)^5 \][/tex]
[tex]\[ f'(1.4) = -24 \cdot (6 - 5.6)^5 \][/tex]
[tex]\[ f'(1.4) = -24 \cdot 0.4^5 \][/tex]

Since [tex]\( 0.4^5 \)[/tex] is a positive number and is multiplied by -24, [tex]\( f'(1.4) \)[/tex] will be negative. Thus, the function is decreasing for [tex]\( x < 1.5 \)[/tex].

- For [tex]\( x > 1.5 \)[/tex]:

Select a point slightly more than 1.5, say [tex]\( x = 1.6 \)[/tex]:

[tex]\[ f'(x) = -24 \cdot (6 - 4 \cdot 1.6)^5 \][/tex]
[tex]\[ f'(1.6) = -24 \cdot (6 - 6.4)^5 \][/tex]
[tex]\[ f'(1.6) = -24 \cdot (-0.4)^5 \][/tex]

Since [tex]\( (-0.4)^5 \)[/tex] is a negative number and is multiplied by -24, [tex]\( f'(1.6) \)[/tex] will be positive. Thus, the function is increasing for [tex]\( x > 1.5 \)[/tex].

### Conclusion
The critical value of [tex]\( f(x) \)[/tex] occurs at [tex]\( A = 1.5 \)[/tex].

For [tex]\( x < 1.5 \)[/tex], [tex]\( f(x) \)[/tex] is decreasing.
For [tex]\( x > 1.5 \)[/tex], [tex]\( f(x) \)[/tex] is increasing.
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