# For g(t)=t^3 +3; how do you find g(1),g(-5),and g(0)?

Aug 1, 2015

A variable like $t$ is a kind of 'place holder' for any value or expression.

Put $1$ in place of $t$ to find: $g \left(1\right) = 4$

Similarly $g \left(- 5\right) = - 122$ and $g \left(0\right) = 3$

#### Explanation:

Put $\textcolor{red}{1}$ in place of $t$ to find:

$g \left(\textcolor{red}{1}\right) = {\left(\textcolor{red}{1}\right)}^{3} + 3 = 1 + 3 = 4$

Putting $\textcolor{red}{- 5}$ in place of $t$ we find:

$g \left(\textcolor{red}{- 5}\right) = {\left(\textcolor{red}{- 5}\right)}^{3} + 3 = - 125 + 3 = - 122$

Putting $\textcolor{red}{0}$ in place of $t$ we find:

$g \left(\textcolor{red}{0}\right) = {\left(\textcolor{red}{0}\right)}^{3} + 3 = 0 + 3 = 3$

You can put expressions in place of $t$ too.

For example,

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

$= \left(8 {x}^{3} + 12 {x}^{2} + 6 x + 1\right) + 3$

$= 8 {x}^{3} + 12 {x}^{2} + 6 x + 4$