Question

question 14 - 1 point
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)=4+\frac{5x^{3}}{3}+\frac{15x^{2}}{2}
provide your answer below:

• local maxima occur at ( x = )
• local minima occur at ( x = )

Explanation:

Step1: Find the first - derivative

Using the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $f(x)=4+\frac{5x^{3}}{3}+\frac{15x^{2}}{2}$, we have $f'(x)=0 + 5x^{2}+15x=5x(x + 3)$.

Step2: Find the critical points

Set $f'(x)=0$. So $5x(x + 3)=0$. Solving this equation gives $x = 0$ or $x=-3$.

Step3: Create a sign - chart for $f'(x)$

Choose test points in the intervals $(-\infty,-3)$, $(-3,0)$ and $(0,\infty)$. Let's choose $x=-4$, $x = - 1$ and $x = 1$.
For $x=-4$, $f'(-4)=5\times(-4)\times(-4 + 3)=20>0$.
For $x=-1$, $f'(-1)=5\times(-1)\times(-1 + 3)=-10<0$.
For $x = 1$, $f'(1)=5\times1\times(1 + 3)=20>0$.

Step4: Determine local extrema

Since $f'(x)$ changes sign from positive to negative at $x=-3$, $f(x)$ has a local maximum at $x=-3$.
Since $f'(x)$ changes sign from negative to positive at $x = 0$, $f(x)$ has a local minimum at $x = 0$.

Answer:

Local maxima occur at $x=-3$.
Local minima occur at $x = 0$.