# For the function g(x)=x^3+x^2-6x, how do you find g(1), g(-1)?

Jun 14, 2017

See a solution process below:

#### Explanation:

To find $g \left(1\right)$ substitute $\textcolor{red}{1}$ for each occurrence of $\textcolor{red}{x}$ in $g \left(x\right)$:

$g \left(\textcolor{red}{x}\right) = {\textcolor{red}{x}}^{3} + {\textcolor{red}{x}}^{2} - 6 \textcolor{red}{x}$ becomes:

$g \left(\textcolor{red}{1}\right) = {\textcolor{red}{1}}^{3} + {\textcolor{red}{1}}^{2} - \left(6 \cdot \textcolor{red}{1}\right)$

$g \left(\textcolor{red}{1}\right) = 1 + 1 - 6$

$g \left(\textcolor{red}{1}\right) = - 4$

To find $g \left(- 1\right)$ substitute $\textcolor{red}{- 1}$ for each occurrence of $\textcolor{red}{x}$ in $g \left(x\right)$:

$g \left(\textcolor{red}{x}\right) = {\textcolor{red}{x}}^{3} + {\textcolor{red}{x}}^{2} - 6 \textcolor{red}{x}$ becomes:

$g \left(\textcolor{red}{- 1}\right) = {\textcolor{red}{- 1}}^{3} + {\textcolor{red}{- 1}}^{2} - \left(6 \cdot \textcolor{red}{- 1}\right)$

$g \left(\textcolor{red}{1}\right) = - 1 + 1 + 6$

$g \left(\textcolor{red}{1}\right) = 6$