consider the following function. g(x)=x³ + 3x² - 6 on -3,1 find the derivative of the function. g(x)= find…

consider the following function. g(x)=x³ + 3x² - 6 on -3,1 find the derivative of the function. g(x)= find any critical numbers of the function. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.) x = find the absolute maximum value and the absolute minimum value, if any, of the function. (if an answer does not exist, enter dne.) maximum minimum need help? read it

consider the following function. g(x)=x³ + 3x² - 6 on -3,1 find the derivative of the function. g(x)= find any critical numbers of the function. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.) x = find the absolute maximum value and the absolute minimum value, if any, of the function. (if an answer does not exist, enter dne.) maximum minimum need help? read it

Answer

Explanation:

Step1: Find the derivative

Use the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$. $g(x)=x^{3}+3x^{2}-6$, so $g'(x)=3x^{2}+6x$.

Step2: Find critical numbers

Set $g'(x) = 0$. $3x^{2}+6x=0$. Factor out $3x$: $3x(x + 2)=0$. Then $3x=0$ or $x + 2=0$. Solving gives $x=0$ or $x=-2$.

Step3: Evaluate the function at critical numbers and endpoints

Evaluate $g(x)$ at $x=-3,x=-2,x = 0,x = 1$. $g(-3)=(-3)^{3}+3(-3)^{2}-6=-27 + 27-6=-6$. $g(-2)=(-2)^{3}+3(-2)^{2}-6=-8 + 12-6=-2$. $g(0)=0^{3}+3(0)^{2}-6=-6$. $g(1)=1^{3}+3(1)^{2}-6=1 + 3-6=-2$.

Answer:

$g'(x)=3x^{2}+6x$ $x=-2,0$ maximum: $-2$ minimum: $-6$