Question

Find (a) \(f + g\), (b) \(f - g\), (c) \(fg\), and (d) \(\frac{f}{g}\) and state their domains.

33. \(f\left( x \right) = \sqrt {25 - {x^2}} ,g\left( x \right) = \sqrt {x + {\bf{1}}} \)

Short answer

a) \(\left( {f + g} \right)\left( x \right) = \sqrt {25 - {x^2}} + \sqrt {x + 1} \). The domain of \(f + g\) is \(\left( { - 1,5} \right)\).

b) \(\left( {f - g} \right)\left( x \right) = \sqrt {25 - {x^2}} - \sqrt {x + 1} \). The domain of \(f - g\) is \(\left( { - 1,5} \right)\).

c) \(\left( {fg} \right)\left( x \right) = \sqrt { - {x^3} - {x^2} + 25x + 25} \). The domain of \(fg\) is \(\left( { - 1,5} \right)\).

d) \(\left( {\frac{f}{g}} \right)\left( x \right) = \sqrt {\frac{{25 - {x^2}}}{{x + 1}}} \). The domain of \(\frac{f}{g}\) is \(\left( { - 1,5} \right)\).

Step-by-step solution

Combination of functions

Consider two functions \(f\) and \(g\), the sum, difference, product,andquotient functionare defined as shown below:

\(\left( {f + g} \right)\left( x \right) = f\left( x \right) + g\left( x \right)\)

\(\left( {f - g} \right)\left( x \right) = f\left( x \right) - g\left( x \right)\)

\(\left( {fg} \right)\left( x \right) = f\left( x \right)g\left( x \right)\)

\(\left( {\frac{f}{g}} \right)\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\)

Determine the domain of the function

The function \(f\left( x \right) = \sqrt {25 - {x^2}} \) is defined only if \(25 - {x^2} \ge 0\). So, it can be written as shown below:

\(\begin{aligned}25 - {x^2} &\ge 0\\{x^2} &\le 25\\ - 5 &\le x \le 5\end{aligned}\)

Thus, the domain of the function \(f\) is \(\left( { - 5,5} \right)\).

The function \(g\left( x \right) = \sqrt {x + 1} \) is defined only when \(x + 1 \ge 0\). So, it can be written as shown below:

\(\begin{aligned}x + 1 &\ge 0\\x &\ge - 1\end{aligned}\)

The domain of the function \(g\) is \(\left( { - 1,\infty } \right)\).

Determine \(f + g,{\rm{ }}f - g,{\rm{ }}fg,\,\,{\mathop{\rm and}\nolimits} \,\,\,\frac{f}{g}\) and obtain their domain

a)

Obtain the function \(\left( {f + g} \right)\left( x \right)\):

\(\left( {f + g} \right)\left( x \right) = \sqrt {25 - {x^2}} + \sqrt {x + 1} \)

Intersect the domain of the two functions as shown below:

\(\left( { - 5,5} \right) \cap \left( { - 1,\infty } \right) = \left( { - 1,5} \right)\)

The domain of the function \(f + g\) is \(\left( { - 1,5} \right)\).

b)

Obtain the function \(\left( {f - g} \right)\left( x \right)\):

\(\left( {f - g} \right)\left( x \right) = \sqrt {25 - {x^2}} - \sqrt {x + 1} \)

Intersect the domain of the two functions as shown below:

\(\left( { - 5,5} \right) \cap \left( { - 1,\infty } \right) = \left( { - 1,5} \right)\)

The domain of \(f - g\) is \(\left( { - 1,5} \right)\).

c)

Obtain the function \(\left( {fg} \right)\left( x \right)\):

\(\begin{aligned}\left( {fg} \right)\left( x \right) &= \sqrt {25 - {x^2}} \cdot \sqrt {x + 1} \\ &= \sqrt { - {x^3} - {x^2} + 25x + 25} \end{aligned}\)

Intersect the domain of the two functions as shown below:

\(\left( { - 5,5} \right) \cap \left( { - 1,\infty } \right) = \left( { - 1,5} \right)\)

The domain of the function \(fg\) is \(\left( { - 1,5} \right)\).

d)

Obtain the function \(\left( {f/g} \right)\left( x \right)\):

\(\begin{aligned}\left( {\frac{f}{g}} \right)\left( x \right) &= \frac{{\sqrt {25 - {x^2}} }}{{\sqrt {x + 1} }}\\ &= \sqrt {\frac{{25 - {x^2}}}{{x + 1}}} \end{aligned}\)

In addition to any previous restrictions, we should have \(x + 1 \ne 0\).

Intersect the domain of the two functions as shown below:

\(\left( { - 5,5} \right) \cap \left( { - 1,\infty } \right) = \left( { - 1,5} \right)\)

The domain of the function \(\frac{f}{g}\) is \(\left( { - 1,5} \right)\).