Question

find the local maxima and minima for the following function: f(x)=-x^3 + 3x^2+1. for any answers that contain a decimal, round to the nearest tenth! please do not put any spaces in your answers!

Explanation:

Step1: Find the derivative

The derivative of $f(x)=-x^{3}+3x^{2}+1$ is $f'(x)=-3x^{2}+6x$.

Step2: Set the derivative equal to zero

Set $-3x^{2}+6x = 0$. Factor out $-3x$ to get $-3x(x - 2)=0$. Then $x = 0$ or $x=2$.

Step3: Use the second - derivative test

The second - derivative $f''(x)=-6x + 6$.
When $x = 0$, $f''(0)=6>0$, so $f(x)$ has a local minimum at $x = 0$. Substitute $x = 0$ into $f(x)$: $f(0)=-0^{3}+3\times0^{2}+1 = 1$.
When $x = 2$, $f''(2)=-6\times2 + 6=-6<0$, so $f(x)$ has a local maximum at $x = 2$. Substitute $x = 2$ into $f(x)$: $f(2)=-2^{3}+3\times2^{2}+1=-8 + 12+1 = 5$.

Answer:

Local maxima: $5$; Local minima: $1$