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 $f'(x)=\frac{1}{9}\times3x^{2}-\frac{1}{3}\times2x-1=\frac{1}{3}x^{2}-\frac{2}{3}x - 1$.

Step2: Set the derivative equal to zero

Set $f'(x) = 0$, so $\frac{1}{3}x^{2}-\frac{2}{3}x - 1=0$. Multiply through by 3 to get $x^{2}-2x - 3 = 0$. Factor the quadratic equation: $(x - 3)(x+1)=0$. Solving for $x$, we get $x=-1$ and $x = 3$.

Step3: Use the first - derivative test

Choose test points in the intervals $(-\infty,-1)$, $(-1,3)$ and $(3,\infty)$. For the interval $(-\infty,-1)$, let $x=-2$. Then $f'(-2)=\frac{1}{3}\times(-2)^{2}-\frac{2}{3}\times(-2)-1=\frac{4}{3}+\frac{4}{3}-1=\frac{4 + 4-3}{3}=\frac{5}{3}>0$, so the function is increasing on $(-\infty,-1)$. For the interval $(-1,3)$, let $x = 0$. Then $f'(0)=\frac{1}{3}\times0^{2}-\frac{2}{3}\times0-1=-1<0$, so the function is decreasing on $(-1,3)$. For the interval $(3,\infty)$, let $x = 4$. Then $f'(4)=\frac{1}{3}\times4^{2}-\frac{2}{3}\times4-1=\frac{16}{3}-\frac{8}{3}-1=\frac{16 - 8-3}{3}=\frac{5}{3}>0$, so the function is increasing on $(3,\infty)$. Since the function changes from increasing to decreasing at $x=-1$, $f(-1)=\frac{1}{9}\times(-1)^{3}-\frac{1}{3}\times(-1)^{2}-(-1)+3=-\frac{1}{9}-\frac{1}{3}+1 + 3=-\frac{1}{9}-\frac{3}{9}+\frac{9}{9}+\frac{27}{9}=\frac{-1-3 + 9+27}{9}=\frac{32}{9}$ is a relative maximum. Since the function changes from decreasing to increasing at $x = 3$, $f(3)=\frac{1}{9}\times3^{3}-\frac{1}{3}\times3^{2}-3 + 3=1 - 3-3 + 3=-2$ is a relative minimum.

Answer:

Relative maximum: $f(-1)=\frac{32}{9}$; Relative minimum: $f(3)=-2$; Increasing intervals: $(-\infty,-1)$ and $(3,\infty)$; Decreasing interval: $(-1,3)$