First, let's find the derivative:
[tex]\begin{gathered} F^{\prime}(x)=3x^{3-1}-3(1) \\ F^{\prime}(x)=3x^2-3 \end{gathered}[/tex]Let's find the critical values:
[tex]\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}[/tex]Since we need to find the absolute extrema for the interval [0,3], we only take x = 1.
Evaluate:
[tex]\begin{gathered} x=0 \\ F(0)=0 \\ ---------- \\ x=1 \\ F(1)=1-3=-2 \\ ------- \\ x=3 \\ F(3)=27-9=18 \end{gathered}[/tex]Since F(1)
