Question

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 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.
$f(x)=3 - x^{3}-3x^{2}$
local maxima occur at x =
local minima occur at x =
provide your answer below:

Explanation:

Step1: Find the first - derivative of the function

Given \(f(x)=3 - x^{3}-3x^{2}\), using the power rule \((x^n)^\prime=nx^{n - 1}\), we have \(f^\prime(x)=-3x^{2}-6x=-3x(x + 2)\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(-3x(x + 2)=0\). Solving this equation gives \(x = 0\) and \(x=-2\).

Step3: Use the First - Derivative Test

Choose test points in the intervals \((-\infty,-2)\), \((-2,0)\) and \((0,\infty)\).
For the interval \((-\infty,-2)\), let's choose \(x=-3\). Then \(f^\prime(-3)=-3\times(-3)\times(-3 + 2)=-9<0\), so the function is decreasing on \((-\infty,-2)\).
For the interval \((-2,0)\), let's choose \(x=-1\). Then \(f^\prime(-1)=-3\times(-1)\times(-1 + 2)=3>0\), so the function is increasing on \((-2,0)\).
For the interval \((0,\infty)\), let's choose \(x = 1\). Then \(f^\prime(1)=-3\times1\times(1 + 2)=-9<0\), so the function is decreasing on \((0,\infty)\).

Since the function changes from decreasing to increasing at \(x=-2\), there is a local minimum at \(x=-2\). Since the function changes from increasing to decreasing at \(x = 0\), there is a local maximum at \(x = 0\).

Answer:

Local maxima occur at \(x = 0\).
Local minima occur at \(x=-2\).