Question

Find the domain of the function.

40. \(f\left( x \right) = \frac{{{x^2} + 1}}{{{x^2} + 4x - 21}}\)

Short answer

The domain of the function is \(\left\{ {\left. {x \in \mathbb{R}} \right|\,\,x \ne \, - 7,\,3} \right\}\) or \(\left( { - \infty , - 7} \right) \cup \left( { - 7,\,3} \right) \cup \left( {3,\,\infty } \right)\).

Step-by-step solution

Analyze the given function

The function \(f\left( x \right) = \frac{{{x^2} + 1}}{{{x^2} + 4x - 21}}\) becomes undefined only when its denominator attains the value 0, that is, when \({x^2} + 4x - 21 = 0\).

\({x^2} + 4x - 21 = 0\)

Solve the quadratic equation by factorizing the polynomial as shown below:

\(\begin{aligned}{x^2} + 4x - 21 &= 0\\\left( {x + 7} \right)\left( {x - 3} \right) &= 0\\x &= - 7,\,\,3\end{aligned}\)

On solving this equation, the values are \(x = - 7,\,\,3\). This implies that the denominator becomes 0 for \(x = - 7,\,\,3\) , and the function \(f\left( x \right) = \frac{{{x^2} + 1}}{{{x^2} + 4x - 21}}\) becomes undefined.

Draw a conclusion

The domain of the function \(f\left( x \right) = \frac{{{x^2} + 1}}{{{x^2} + 4x - 21}}\) must exclude the values \(x = - 7,\,\,3\) as it is undefined for these values, whereas for all the other values function attains a finite value.

So, the domain of the function can be written as \(\left\{ {\left. {x \in \mathbb{R}} \right|\,\,x \ne \, - 7,\,3} \right\}\) or \(\left( { - \infty , - 7} \right) \cup \left( { - 7,\,3} \right) \cup \left( {3,\,\infty } \right)\).