Precalculus

Dr. Lalitha Subramanian
About the Author

Aksharam

Chapter 1: Functions and Graphs

1.3: Operations on Functions.

Given two functions $y=f(x)$ and $y=g(x)$, we can evaluate any expression involving addition, subtraction, multiplication, division, or composition of functions. Rules for doing this are stated below:

Example 1:

Let $f(x)=3x+2$ and $g(x)=\frac{3x-2}{x}$, find and simplify the following:
(a) $\quad (fg)(x)$
(b)$\quad (f-g)(x)$

Solution:(a) \begin{eqnarray} (fg)(x) &=& (3x+2)\left(\frac{3x-2}{x}\right)\\ &=& \frac{(3x+2)(3x-2)}{x}\\ &=& \frac{9x^{2}-4}{x}\\ \end{eqnarray}
(b) \begin{eqnarray} (f-g)(x) &=& (3x+2) - \frac{3x-2}{x}\\ &=& \frac{x(3x+2)-(3x-2)}{x}\\ &=& \frac{3x^{2}+2x-3x+2}{x}\\ &=& \frac{3x^{2}-x+2}{x}\\ \end{eqnarray}

Example 2:

Given two functions $f(x)=\sqrt{x+2}$ and $g(x)=x^2$, find and simplify
(a) $\quad (f \circ g)(x)]$
(b) $\quad \left(\frac{f}{g}\right)(x)$

Solution:(a) \begin{eqnarray} (f \circ g)(x) &=& f[g(x)]\\ &=& f(x^2)\\ &=& \sqrt{x^2+2}\\ \end{eqnarray}
(b) \begin{eqnarray} \left(\frac{f}{g}\right)(x) &=& \frac{\sqrt{x+2}}{x^2} \\ \end{eqnarray}.

Example 3:

Let $f(x)=\sqrt{x+3}$ and $g(x)=2x-1$. Evaluate
(a) $\quad (fg)\left(\frac{1}{2}\right)$
(b) $\quad \left(\frac{f}{g}\right)(-2)$.

Solution:(a) \begin{eqnarray} (fg)\left(\frac{1}{2}\right) &=& f\left(\frac{1}{2}\right) \times g \left(\frac{1}{2}\right)\\ &=& \sqrt{\left(\frac{1}{2}+3\right)} \times \left(2 \times \frac{1}{2}-1 \right)\\ &=& \left(\sqrt{\frac{7}{2}}\right) \times 0 \\ &=& 0\\ \end{eqnarray}
(b) \begin{eqnarray} \left(\frac{f}{g}\right)(-2) &=& \left(\frac{f(-2)}{g(-2)}\right)\\ &=& \left(\frac{\sqrt{-2+3}}{-4-1}\right)\\ &=& \left(\frac{1}{-5}\right)\\ &=& -\left(\frac{1}{5}\right)\\ \end{eqnarray}

Example 4:

Let $f(x)=2x+1$ and $g(x)=x-2$. Find and simplify $(f \circ g)(x)$.

Solution:\begin{eqnarray} (f \circ g)(x) &=& f[g(x)]\\ &=& f[x-2]\\ &=& 2(x-2)+1\\ &=& 2x-4+1\\ &=& 2x-3\\ \end{eqnarray}

Example 5:

Let $f(x)=x^2+4$ and $g(x)=\sqrt{x+1}$. Evaluate
(a) $\quad (f\circ g)(3)$.
(b) $\quad (g\circ f)(-2)$.

Solution:(a) \begin{eqnarray} (f\circ g)(3) &=& f[g(3)] \\ &=& f[\sqrt{4}]\\ &=& f[2]\\ &=& 2^2+4\\ &=& 8\\ \end{eqnarray}
(b) \begin{eqnarray} (g \circ f)(-2) &=& g[f(-2)]\\ &=& g[8]\\ &=& \sqrt{9}\\ &=& \pm (3)\\ \end{eqnarray}

Example 6:

Let $f(x)=\frac{1}{x-1}$ and $g(x)=\sqrt{x}$. Determine
(a) $\quad (f\circ g)(x)$.
(b) $\quad (g\circ f)(x)$.

Solution:(a) \begin{eqnarray} (f\circ g)(x) &=& f[g(x)] \\ &=& f[\sqrt{x}]\\ &=& \frac{1}{\sqrt{x}-1}\\ \end{eqnarray}
(b) \begin{eqnarray} (g \circ f)(x) &=& g[f(x)]\\ &=& g[\frac{1}{x-1}]\\ &=& \sqrt{\frac{1}{x-1}}\\ &=& \frac{1}{\sqrt{x-1}}\\ \end{eqnarray}

From the above example, we note that $(f \circ g)(x) \neq (g \circ f)(x)$. Generally, composition of two functions is not commutative. That is $(f \circ g)(x) \neq (g \circ f)(x)$.

For the composite function $(f\circ g)(x)$ to be meaningful, $g(x)$ must be in the domain of $f(x)$. To find the domain of $(f\circ g)(x)$, we need to consider the domain of $f(x)$ and the domain of $g(x)$.

Example 7:

Let $f(x)=x^2+2$ and $g(x)=\sqrt{x}$. Determine the following composite functions and domains:
(a) $\quad (f\circ g)(x)$.
(b) $\quad (g\circ f)(x)$.

Solution:
(a) \begin{eqnarray} (f\circ g)(x) &=& f[g(x)] \\ &=& f[\sqrt{x}]\\ &=& x+2\\ \end{eqnarray}
Domain of $f(x)$ is all real numbers, but the domain of $g(x)$ is are non-negative real numbers. This makes the domain of $f[g(x)]$ as all non-negative real numbers. So, domain of $(f\circ g)(x)$ is $[0, \infty)$.

(b) \begin{eqnarray} (g \circ f)(x) &=& g[f(x)]\\ &=& g[x^2+2]\\ &=& \sqrt{x^{2}+2}\\ \end{eqnarray}
In this case, domain of the inside function $f(x)=x^2+2$ is all real numbers, whereas the domain of the outside function $g(x)=\sqrt{x}$ is all non-negative real numbers. We note that the domain of $(g \circ f)(x)$ is $(-\infty, \infty)$.

One important operation of function involves finding the difference quotient of a function.
For any given function $y=f(x)$, the expression $\frac{f(x+h)-f(x)}{h}$ is called the difference quotient of the function. This difference quotient is of great relevance in calculus.
We can split the process of finding the difference quotient into 3 steps:

Example 8:

Find the simplify the difference quotient for the function $f(x)=x^2+x-3$

Solution:
\begin{eqnarray} f(x+h) &=& (x+h)^2 +(x+h)-3\\ &=& x^2+2hx+h^2+x+h-3\\ f(x+h)-f(x) &=& (x^2+2hx+h^2+x+h-3)-(x^2+x-3)\\ &=& 2hx+h^2+h\\ \frac{f(x+h)-f(x)}{h} &=& \frac{h(2x+h+1)}{h}\\ &=& 2x+h+1\\ \end{eqnarray}

Example 9:

Find and simplify the difference quotient for the function $f(x) = \sqrt{x}$

Solution: For this function, step 1 is not needed. So, we shall start with step 2: \begin{eqnarray} f(x+h)-f(x) &=& \sqrt{x+h} - \sqrt{x}\\ \frac{f(x+h)-f(x)}{h} &=& \frac{\sqrt{x+h}-\sqrt{x}}{h}\\ \end{eqnarray} In order to simplify this expression so that the $h$ in the denominator can be canceled, we multiply and divide the expression by the conjugate of the numerator: \begin{eqnarray} \frac{f(x+h)-f(x)}{h} &=& \frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}\\ &=& \frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\\ &=& \frac{h}{h(\sqrt{x+h}+\sqrt{x})}\\ &=& \frac{1}{\sqrt{x+h} + \sqrt{x}}\\ \end{eqnarray}

Example 10:

Find and simplify the difference quotient for the function $g(x) = \frac{2}{x+1}$

Solution:\begin{eqnarray} g(x+h) &=& \frac{2}{x+h+1}\\ g(x+h)-g(x) &=& \frac{2}{x+h+1}-\frac{2}{x+1}\\ &=& \frac{2(x+1)-2(x+h+1)}{(x+h+1)(x+1)}\\ &=& \frac{2x+2-2x-2h-2}{(x+h+1)(x+1)}\\ &=& \frac{-2h}{(x+h+1)(x+1)}\\ \frac{g(x+h)-g(x)}{h} &=& \frac{-2j}{h(x+h+1)(x+1)}\\ &=& \frac{-2}{(x+h+1)(x+1)}\\ \end{eqnarray}