throughout which interval is $f(x)=-x^{3}+2x^{2}+4x - 2$ increasing?\n$(-\\infty,-3$\n$-3,0$\n$0,2)$\n$(2,\\i…

Explanation:

Step1: Find the derivative

The derivative of $f(x)=-x^{3}+2x^{2}+4x - 2$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=-3x^{2}+4x + 4$.

Step2: Set the derivative greater than 0

We want to find where $f'(x)>0$, so we solve the inequality $-3x^{2}+4x + 4>0$. First, rewrite it as $3x^{2}-4x - 4<0$.

Step3: Factor the quadratic

Factor $3x^{2}-4x - 4$: $3x^{2}-4x - 4=(3x + 2)(x - 2)<0$.

Step4: Find the roots

Set each factor equal to 0: $3x+2 = 0$ gives $x=-\frac{2}{3}$ and $x - 2=0$ gives $x = 2$.

Step5: Test intervals

The intervals to test are $(-\infty,-\frac{2}{3})$, $(-\frac{2}{3},2)$ and $(2,\infty)$. For $x=-1$ in $(-\infty,-\frac{2}{3})$, $(3(-1)+2)(-1 - 2)=(-1)(-3)=3>0$. For $x = 0$ in $(-\frac{2}{3},2)$, $(3(0)+2)(0 - 2)=(2)(-2)=-4<0$. For $x = 3$ in $(2,\infty)$, $(3(3)+2)(3 - 2)=(11)(1)=11>0$. So the function $f(x)$ is increasing on the interval $(-\frac{2}{3},2)$. Among the given options, the closest correct interval is $[0,2)$.

Answer:

C. $[0,2)$