Plot the following points for the table,
x = 0, 1, 4, 8, 12, 16
[tex]\begin{gathered} f(x)=4-x^{\frac{3}{2}} \\ \\ f(0)=4-0^{\frac{3}{2}} \\ f(0)=4 \\ \text{First point }(0,4) \\ \\ f(1)=4-1^{\frac{3}{2}} \\ f(1)=4-1 \\ f(1)=3 \\ \text{Second point }(1,3) \\ \\ f(4)=4-4^{\frac{3}{2}} \\ f(4)=4-\sqrt[]{4^3} \\ f(4)=4-\sqrt[]{64} \\ f(4)=4-8 \\ f(4)=-4 \\ \text{Third point }(4,-4)_{} \\ \\ f(8)=4-8^{\frac{3}{2}} \\ f(8)=4-\sqrt[]{8^3} \\ f(8)\approx-18.62 \\ \text{Fourth point }(8,-18.63) \\ \\ f(12)=4-12^{\frac{3}{2}} \\ f(12)=4-\sqrt[]{12^3} \\ f(12)\approx-37.57 \\ \text{Fifth point }(12,-37.57) \\ \\ f(16)=4-16^{\frac{3}{2}} \\ f(16)=4-\sqrt[]{16^3} \\ f(16)=-60 \\ \text{Sixth point }(16,-60) \end{gathered}[/tex]
The table therefore is
The graph f(x) is
