Let's solve the problem step-by-step.
We start with the function:
[tex]\[ f(x) = (4 + 3x)^4 \][/tex]
### 1. Find the Derivative
To find critical points, we need to determine the derivative of [tex]\( f(x) \)[/tex].
[tex]\[ f'(x) = \frac{d}{dx}[(4 + 3x)^4] \][/tex]
Using the chain rule:
[tex]\[ f'(x) = 4(4 + 3x)^3 \cdot \frac{d}{dx}(4 + 3x) \][/tex]
[tex]\[ f'(x) = 4(4 + 3x)^3 \cdot 3 \][/tex]
[tex]\[ f'(x) = 12(4 + 3x)^3 \][/tex]
### 2. Find Critical Points
Set the derivative equal to zero to find the critical points.
[tex]\[ 12(4 + 3x)^3 = 0 \][/tex]
Divide both sides by 12:
[tex]\[ (4 + 3x)^3 = 0 \][/tex]
Taking the cube root of both sides:
[tex]\[ 4 + 3x = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 3x = -4 \][/tex]
[tex]\[ x = -\frac{4}{3} \][/tex]
Thus, the critical value [tex]\( A \)[/tex] is:
[tex]\[ A = -\frac{4}{3} \][/tex]
[tex]\[ A \approx -1.3333333333333333 \][/tex]
### 3. Determine the Behavior Around the Critical Point
To determine the behavior of [tex]\( f(x) \)[/tex] around [tex]\( A \)[/tex], we analyze [tex]\( f'(x) \)[/tex] for values slightly less than and greater than [tex]\( -\frac{4}{3} \)[/tex].
#### For [tex]\( x < A \)[/tex]:
Choose a value slightly less than [tex]\( -\frac{4}{3} \)[/tex]. For illustration, let's consider [tex]\( x \approx -1.333 - 1 = -2.333 \)[/tex].
[tex]\[ f'(-2.333) \approx 12(4 + 3(-2.333))^3 \][/tex]
[tex]\[ = 12(4 - 6.999)^3 \][/tex]
[tex]\[ = 12(-2.999)^3 \][/tex]
[tex]\[ = 12 \cdot (-27.027) \][/tex]
[tex]\[ \approx -324.324 \][/tex]
Since [tex]\( f'(-2.333) < 0 \)[/tex], the function is decreasing for [tex]\( x < -\frac{4}{3} \)[/tex].
#### For [tex]\( x > A \)[/tex]:
Choose a value slightly greater than [tex]\( -\frac{4}{3} \)[/tex]. For illustration, let's consider [tex]\( x \approx -1.333 + 1 = -0.333 \)[/tex].
[tex]\[ f'(-0.333) \approx 12(4 + 3(-0.333))^3 \][/tex]
[tex]\[ = 12(4 - 0.999)^3 \][/tex]
[tex]\[ = 12(3.001)^3 \][/tex]
[tex]\[ = 12 \cdot 27.027 \][/tex]
[tex]\[ \approx 324.324 \][/tex]
Since [tex]\( f'(-0.333) > 0 \)[/tex], the function is increasing for [tex]\( x > -\frac{4}{3} \)[/tex].
### Summary
- The critical value is at [tex]\( A = -\frac{4}{3} \approx -1.3333333333333333 \)[/tex].
- For [tex]\( x < -\frac{4}{3} \)[/tex], [tex]\( f(x) \)[/tex] is decreasing.
- For [tex]\( x > -\frac{4}{3} \)[/tex], [tex]\( f(x) \)[/tex] is increasing.
