Chapter 2, Problem 4CT

To determine

The domain of function $h(x)=\frac{x-4}{x^2+5x-36}$, and evaluate function at $x=-1$ .

Answer to Problem 4CT

Solution:

Domain of the function is $\left(-\infty ,-9\right)\cup \left(-9,4\right)\cup \left(4,\infty \right)$ .

The value of the function $h(-1)=\frac{1}{8}$ .

Explanation of Solution

Given information:

The function is $h(x)=\frac{x-4}{x^2+5x-36}$ .

Domain is set of $x$ values where the function is defined.

The function $h(x)=\frac{x-4}{x^2+5x-36}$ is defined for all values except $x^2+5x-36=0$ .

Function is undefined at zeroes of polynomial $x^2+5x-36=0$ .

Consider $x^2+5x-36=0$ .

Find zeroes of the polynomial.

The product of $9\text{and}-4$ is $-36$, and sum of $9\text{and}-4$ is $5$ .

Therefore, the polynomial becomes $(x+9)(x-4)=0$ .

  $\Rightarrow x+9=0\text{or}x-4=0$ .

  $\Rightarrow x=-9\text{or}x=4$ .

Therefore, exclude the points $x=-9,4$ from domain.

Thus, domain of function in interval notation is $\left(-\infty ,-9\right)\cup \left(-9,4\right)\cup \left(4,\infty \right)$ .

To evaluate function at $x=-1$, substitute $x=-1$ in $h(x)=\frac{x-4}{x^2+5x-36}$ .

Therefore,

  $h(-1)=\frac{-1-4}{{(-1)}^2+5(-1)-36}$ ,

  $\Rightarrow h(-1)=\frac{-5}{1-5-36}=\frac{-5}{-40}$ ,

  $\Rightarrow h(-1)=\frac{-5}{-40}=\frac{1}{8}$ ,

Thus, the value of $h(-1)=\frac{1}{8}$ .