$\begin{cases} f(x)=2x^{4}+x^{3}-11x^{2}+11x-3, \ 2, \frac{3}{2} \end{cases}$

Full solution

\begin{cases} f(x)=2x^{4}+x^{3}-11x^{2}+11x-3, \ 2, \frac{3}{2} \end{cases}

Substitute $x$ with $2$ : To find the value of $f(x)$ when $x=2$ , we need to substitute $x$ with $2$ in the function $f(x)=2x^4+x^3-11x^2+11x-3$ .
Calculation: $f(2)=2(2)^4+(2)^3-11(2)^2+11(2)-3$
$f(2)=2(16)+8-11(4)+22-3$
$f(2)=32+8-44+22-3$
$2$ $0$
$2$ $1$
$2$ $2$
Calculate $f(2)$ : To find the value of $f(x)$ when $x=\frac{3}{2}$ , we need to substitute $x$ with $\frac{3}{2}$ in the function $f(x)=2x^4+x^3-11x^2+11x-3$ . $\newline$ Calculation: $f\left(\frac{3}{2}\right)=2\left(\frac{3}{2}\right)^4+\left(\frac{3}{2}\right)^3-11\left(\frac{3}{2}\right)^2+11\left(\frac{3}{2}\right)-3$ $\newline$ $f\left(\frac{3}{2}\right)=2\left(\frac{81}{16}\right)+\left(\frac{27}{8}\right)-11\left(\frac{9}{4}\right)+\frac{33}{2}-3$ $\newline$ $f\left(\frac{3}{2}\right)=2\left(\frac{81}{16}\right)+\left(\frac{27}{8}\right)-\frac{99}{4}+\frac{33}{2}-3$ $\newline$ To simplify, we find a common denominator, which is $16$ , and continue the calculation. $\newline$ $f\left(\frac{3}{2}\right)=2\left(\frac{81}{16}\right)+\left(\frac{54}{16}\right)-\left(4\cdot\frac{99}{16}\right)+\left(4\cdot\frac{33}{8}\right)-3$ $\newline$ $f(x)$ $0$ $\newline$ $f(x)$ $1$ $\newline$ $f(x)$ $2$ $\newline$ $f(x)$ $3$ $\newline$ $f(x)$ $4$