Questions: Use the First Derivative Test to find the location of all local extrema for the function given below. Enter an exact answer. If there is more than one local maximum or local minimum, write each value of x separated by a comma. If a local maximum or local minimum does not occur on the function, enter ∅ in the appropriate box.
f(x)=(x+4)^3(3x-2)^2
Provide your answer below:
- Local maxima occur at x= .
- Local minima occur at x= .
Transcript text: Use the First Derivative Test to find the location of all local extrema for the function given below. Enter an exact answer. If there is more than one local maximum or local minimum, write each value of $x$ separated by a comma. If a local maximum or local minimum does not occur on the function, enter $\varnothing$ in the appropriate box.
\[
f(x)=(x+4)^{3}(3 x-2)^{2}
\]
Provide your answer below:
- Local maxima occur at $x=$ $\square$ .
- Local minima occur at $x=$ $\square$ .
Solution
Solution Steps
Step 1: Find the derivative of \(f(x)\)
The derivative of \(f(x)\) is \(f'(x) = \left(x + 4\right)^{3} \cdot \left(18 x - 12\right) + 3 \left(x + 4\right)^{2} \left(3 x - 2\right)^{2}\).
Step 2: Solve for critical points
The critical points are found by solving \(f'(x) = 0\). The solutions are -4, -1.2, 0.67, considering the domain Reals.
Step 3: Apply the First Derivative Test
Local maxima are located at -1.2.
Local minima are located at 0.67.
Final Answer:
Local maxima: Local maxima are located at -1.2. Local minima: Local minima are located at 0.67.