Question

Find the absolute extrema of the following function on the indicted interval. F(x)=x^3-3x on [0,3]

Answer 1

`F(x)=x^3-3x`

First, let's find the derivative:

`\begin{gathered} F^(\prime)(x)=3x^(3-1)-3(1) \\ F^(\prime)(x)=3x^2-3 \end{gathered}`

Let's find the critical values:

`\begin{gathered} F^(\prime)(x)=0 \\ 3x^2-3=0 \\ x^2-1=0 \\ x^2=1 \\ so: \\ x=1 \\ or \\ x=-1 \end{gathered}`

Since we need to find the absolute extrema for the interval [0,3], we only take x = 1.

Evaluate:

`\begin{gathered} x=0 \\ F(0)=0 \\ ---------- \\ x=1 \\ F(1)=1-3=-2 \\ ------- \\ x=3 \\ F(3)=27-9=18 \end{gathered}`

Since F(1)