How do you find (f+g)(-2) given f(x)=x^2-1 and g(x)=2x-3 and h(x)=1-4x#?

Apr 10, 2017

See the entire solution process below:

Explanation:

First,

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

We can ignore $h \left(x\right)$ as it is extraneous information to this problem.

To find $\left(f + g\right) \left(- 2\right)$ we need to substitute $\textcolor{red}{- 2}$ for each occurrence of $\textcolor{red}{x}$ in $\left(f + g\right) \left(x\right)$:

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

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

$\left(f + g\right) \left(\textcolor{red}{- 2}\right) = 4 - 1 - 4 - 3$

$\left(f + g\right) \left(\textcolor{red}{- 2}\right) = - 4$