using the graph, determine any relative maxima or minima of the function and the intervals on which the…

Explanation:

Step1: Find the derivative

Differentiate $f(x)=\frac{1}{9}x^{3}-\frac{1}{3}x^{2}-x + 3$ using the power - rule. The derivative $f^\prime(x)=\frac{1}{9}\times3x^{2}-\frac{1}{3}\times2x - 1=\frac{1}{3}x^{2}-\frac{2}{3}x - 1$.

Step2: Set the derivative equal to zero

Set $f^\prime(x)=0$, so $\frac{1}{3}x^{2}-\frac{2}{3}x - 1 = 0$. Multiply through by 3 to get $x^{2}-2x - 3=0$.

Step3: Factor the quadratic equation

Factor $x^{2}-2x - 3=(x - 3)(x+1)=0$.

Step4: Solve for x

Setting each factor equal to zero gives $x - 3=0$ or $x + 1=0$, so $x=-1$ or $x = 3$.

Step5: Use the second - derivative test

Find the second - derivative $f^{\prime\prime}(x)=\frac{2}{3}x-\frac{2}{3}$. Evaluate $f^{\prime\prime}(-1)=\frac{2}{3}\times(-1)-\frac{2}{3}=-\frac{4}{3}<0$, so $f(x)$ has a relative maximum at $x=-1$. Evaluate $f^{\prime\prime}(3)=\frac{2}{3}\times3-\frac{2}{3}=\frac{4}{3}>0$, so $f(x)$ has a relative minimum at $x = 3$.

Step6: Find the relative maximum value

Substitute $x=-1$ into $f(x)$: $f(-1)=\frac{1}{9}\times(-1)^{3}-\frac{1}{3}\times(-1)^{2}-(-1)+3=-\frac{1}{9}-\frac{1}{3}+1 + 3=-\frac{1}{9}-\frac{3}{9}+4=\frac{-1 - 3}{9}+4=\frac{-4}{9}+4=\frac{-4 + 36}{9}=\frac{32}{9}\approx3.56$.

Answer:

A. The relative maximum value(s) $\frac{32}{9}$ occur(s) at $x=-1$