Q. 4 - Precalculus Enhanced with Graphing Utilities

Question

Find the domain of each function and evaluate each function at $x=-1$.

$h\left(x\right)=\frac{x-4}{x^2+5x-36}$

Step-by-Step Solution

Answer

The domain of the function is $\left\{x|x\ne -9,x\ne 4\right\}$

The value at $x=-1$ is $\frac{1}{8}$

Step 1. Find the domain of the function.

$h\left(x\right)=\frac{x-4}{x^2+5x-36}$

For domain, denominator cannot be zero.

So,

$$\begin{gathered} x^2+5x-36\ne 0 \\ {x}^2+9x-4x-36\ne 0 \\ x(x+9)-4(x+9)\ne 0 \\ (x+9)(x-4)\ne 0 \\ x\ne -9,4 \end{gathered}$$

Therefore, the domain of the function is $x$ belongs to all real numbers except $-9,4$. 

Step 2. Find value of h ( x ) at x = - 1

$$\begin{gathered} h\left(x\right)=\frac{x-4}{x^2+5x-36} \\ h(-1)=\frac{-1-4}{{(-1)}^2+5(-1)-36} \\ h(-1)=\frac{-5}{-40} \\ h(-1)=\frac{1}{8} \end{gathered}$$