# 3.1 Functions and function notation  (Page 5/21)

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## Evaluating functions

Given the function $\text{\hspace{0.17em}}h\left(p\right)={p}^{2}+2p,\text{\hspace{0.17em}}$ evaluate $\text{\hspace{0.17em}}h\left(4\right).\text{\hspace{0.17em}}$

To evaluate $\text{\hspace{0.17em}}h\left(4\right),\text{\hspace{0.17em}}$ we substitute the value 4 for the input variable $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ in the given function.

$\begin{array}{ccc}\hfill h\left(p\right)& =& {p}^{2}+2p\hfill \\ \hfill h\left(4\right)& =& {\left(4\right)}^{2}+2\left(4\right)\hfill \\ & =& 16+8\hfill \\ & =& 24\hfill \end{array}$

Therefore, for an input of 4, we have an output of 24.

Given the function $\text{\hspace{0.17em}}g\left(m\right)=\sqrt{m-4},\text{\hspace{0.17em}}$ evaluate $\text{\hspace{0.17em}}g\left(5\right).$

$\text{\hspace{0.17em}}g\left(5\right)=1\text{\hspace{0.17em}}$

## Solving functions

Given the function $\text{\hspace{0.17em}}h\left(p\right)={p}^{2}+2p,\text{\hspace{0.17em}}$ solve for $\text{\hspace{0.17em}}h\left(p\right)=3.$

If $\text{\hspace{0.17em}}\left(p+3\right)\left(p-1\right)=0,\text{\hspace{0.17em}}$ either $\text{\hspace{0.17em}}\left(p+3\right)=0\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\left(p-1\right)=0\text{\hspace{0.17em}}$ (or both of them equal 0). We will set each factor equal to 0 and solve for $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ in each case.

$\begin{array}{cccccc}\hfill \left(p+3\right)& =& 0,\hfill & \hfill \phantom{\rule{0.5em}{0ex}}p& =& -3\hfill \\ \hfill \left(p-1\right)& =& 0,\hfill & \hfill \phantom{\rule{0.5em}{0ex}}p& =& 1\hfill \end{array}$

This gives us two solutions. The output $\text{\hspace{0.17em}}h\left(p\right)=3\text{\hspace{0.17em}}$ when the input is either $\text{\hspace{0.17em}}p=1\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}p=-3.\text{\hspace{0.17em}}$ We can also verify by graphing as in [link] . The graph verifies that $\text{\hspace{0.17em}}h\left(1\right)=h\left(-3\right)=3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\left(4\right)=24.$

Given the function $\text{\hspace{0.17em}}g\left(m\right)=\sqrt{m-4},\text{\hspace{0.17em}}$ solve $\text{\hspace{0.17em}}g\left(m\right)=2.$

$\text{\hspace{0.17em}}m=8\text{\hspace{0.17em}}$

## Evaluating functions expressed in formulas

Some functions are defined by mathematical rules or procedures expressed in equation    form. If it is possible to express the function output with a formula    involving the input quantity, then we can define a function in algebraic form. For example, the equation $\text{\hspace{0.17em}}2n+6p=12\text{\hspace{0.17em}}$ expresses a functional relationship between $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}p.\text{\hspace{0.17em}}$ We can rewrite it to decide if $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is a function of $\text{\hspace{0.17em}}n.\text{\hspace{0.17em}}$

Given a function in equation form, write its algebraic formula.

1. Solve the equation to isolate the output variable on one side of the equal sign, with the other side as an expression that involves only the input variable.
2. Use all the usual algebraic methods for solving equations, such as adding or subtracting the same quantity to or from both sides, or multiplying or dividing both sides of the equation by the same quantity.

## Finding an equation of a function

Express the relationship $\text{\hspace{0.17em}}2n+6p=12\text{\hspace{0.17em}}$ as a function $\text{\hspace{0.17em}}p=f\left(n\right),\text{\hspace{0.17em}}$ if possible.

To express the relationship in this form, we need to be able to write the relationship where $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is a function of $\text{\hspace{0.17em}}n,\text{\hspace{0.17em}}$ which means writing it as $\text{\hspace{0.17em}}p=\left[\text{expression}\text{\hspace{0.17em}}\text{involving}\text{\hspace{0.17em}}n\right].$

Therefore, $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ is written as

$p=f\left(n\right)=2-\frac{1}{3}n$

## Expressing the equation of a circle as a function

Does the equation $\text{\hspace{0.17em}}{x}^{2}+{y}^{2}=1\text{\hspace{0.17em}}$ represent a function with $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ as input and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as output? If so, express the relationship as a function $\text{\hspace{0.17em}}y=f\left(x\right).$

First we subtract $\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ from both sides.

${y}^{2}=1-{x}^{2}$

We now try to solve for $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ in this equation.

We get two outputs corresponding to the same input, so this relationship cannot be represented as a single function $\text{\hspace{0.17em}}y=f\left(x\right).$ If we graph both functions on a graphing calculator, we will get the upper and lower semicircles.

If $\text{\hspace{0.17em}}x-8{y}^{3}=0,\text{\hspace{0.17em}}$ express $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}x.$

$y=f\left(x\right)=\frac{\sqrt[3]{x}}{2}$

Are there relationships expressed by an equation that do represent a function but that still cannot be represented by an algebraic formula?

Yes, this can happen. For example, given the equation $\text{\hspace{0.17em}}x=y+{2}^{y},\text{\hspace{0.17em}}$ if we want to express $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ there is no simple algebraic formula involving only $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ that equals $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ However, each $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ does determine a unique value for $\text{\hspace{0.17em}}y,\text{\hspace{0.17em}}$ and there are mathematical procedures by which $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ can be found to any desired accuracy. In this case, we say that the equation gives an implicit (implied) rule for $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ even though the formula cannot be written explicitly.

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