Question

A function f is defined by $f\left(x\right)=2x^3-3x,xϵR$. What is the value of $\frac{1}{2}f\left(2\right)$?

Answer 1

Given, $f\left(x\right)=2x^3-3x$

To find: $\frac{1}{2}f\left(2\right)$

Therefore, $\frac{1}{2}f\left(2\right)=\frac{1}{2}[2{\left(2\right)}^3-3(2\left)\right]$

$=\frac{1}{2}\left[2\right(8)-6]$

$=\frac{1}{2}\left(10\right)$

$=5$