Chapter 8, Problem 3CR

To determine

The domain of function $f(x)=\sqrt{x^2-3x-4}$

Answer to Problem 3CR

Solution:

The domain of function $f(x)=\sqrt{x^2-3x-4}$ is $\left(-\infty ,-1\right]\cup \left[4,\infty \right)$

Explanation of Solution

Given information:

The function $f(x)=\sqrt{x^2-3x-4}$

Domain of the function is a set of $x$ values, for which the function is defined

The function $f(x)=\sqrt{x^2-3x-4}$ is defined for non-negative values of radical.

That is, $x^2-3x-4\ge 0$

To solve the inequality, find zeroes of the above term.

Consider $x^2-3x-4=0$

  $\Rightarrow \left(x-4\right)\left(x+1\right)=0$

  $\Rightarrow x=4,x=-1$ .

Form three intervals using the above zeroes and check by taking test points.

  $\begin{array}{cccc}\text{Interval}& \left(-\infty ,-1\right)& \left(-1,4\right)& \left(4,\infty \right)\\ \text{Test}\text{point}& -2& 0& 5\\ \text{Inequality}\text{at}\text{test}\text{point}& \begin{array}{l}={(}^{-}-3(-2)-4\\ =6\ge 0\end{array}& \begin{array}{l}={(}^0-3(0)-4\\ =-4\le 0\end{array}& \begin{array}{l}={(}^5-3(5)-4\\ =6\ge 0\end{array}\end{array}$

The inequality holds true in the intervals $\left(-\infty ,-1\right)$ and $\left(4,\infty \right)$

Now, check the inequality at end points $-1,4$

At $x=-1$, ${(-1)}^2-3(-1)-4=1+3-4=0$

Inequality holds at $x=-1$

At $x=4$, ${(4)}^2-3(4)-4=16-12-4=0$

Inequality holds at $x=4$

The inequality holds true in the interval $\left(-\infty ,-1\right]\cup \left[4,\infty \right)$

Domain of the function $f(x)=\sqrt{x^2-3x-4}$ is $\left(-\infty ,-1\right]\cup \left[4,\infty \right)$