2. given: f(x)=\\frac{1}{3}x^{3}+x^{2}-3x - 3 a) use the fdt (find the critical, relative maximum and…

Explanation:

Step1: Find the derivative

The derivative of $f(x)=\frac{1}{3}x^{3}+x^{2}-3x - 3$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $f^\prime(x)=x^{2}+2x - 3$.

Step2: Find the critical numbers

Set $f^\prime(x)=0$. So $x^{2}+2x - 3 = 0$. Factor the quadratic equation: $(x + 3)(x - 1)=0$. Solving for $x$ gives $x=-3$ and $x = 1$. These are the critical numbers.

Step3: Determine the intervals of increase and decrease

Choose test points in the intervals $(-\infty,-3)$, $(-3,1)$ and $(1,\infty)$. For the interval $(-\infty,-3)$, let $x=-4$. Then $f^\prime(-4)=(-4)^{2}+2(-4)-3=16-8 - 3=5>0$, so the function is increasing on $(-\infty,-3)$. For the interval $(-3,1)$, let $x = 0$. Then $f^\prime(0)=0^{2}+2(0)-3=-3<0$, so the function is decreasing on $(-3,1)$. For the interval $(1,\infty)$, let $x = 2$. Then $f^\prime(2)=2^{2}+2(2)-3=4 + 4-3=5>0$, so the function is increasing on $(1,\infty)$.

Step4: Find the relative extrema

Since the function changes from increasing to decreasing at $x=-3$, $f(-3)=\frac{1}{3}(-3)^{3}+(-3)^{2}-3(-3)-3=-9 + 9+9 - 3=6$ is a relative maximum. Since the function changes from decreasing to increasing at $x = 1$, $f(1)=\frac{1}{3}(1)^{3}+(1)^{2}-3(1)-3=\frac{1}{3}+1-3 - 3=\frac{1 + 3-9 - 9}{3}=-\frac{14}{3}$ is a relative minimum.

Answer:

Critical Numbers: $x=-3,x = 1$ Increasing: $(-\infty,-3)\cup(1,\infty)$ Decreasing: $(-3,1)$ Relative Maximum: $f(-3)=6$ Relative Minimum: $f(1)=-\frac{14}{3}$