1.1 Review of functions (Page 7/28)

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For $f (x) = x^{2} + 3$ and $g (x) = 2 x - 5,$ find $(f / g) (x)$ and state its domain.

$(\frac{f}{g}) (x) = \frac{x^{2} + 3}{2 x - 5} .$ The domain is ${x | x \neq \frac{5}{2}} .$

Function composition

When we compose functions, we take a function of a function. For example, suppose the temperature $T$ on a given day is described as a function of time $t$ (measured in hours after midnight) as in [link] . Suppose the cost $C,$ to heat or cool a building for 1 hour, can be described as a function of the temperature $T .$ Combining these two functions, we can describe the cost of heating or cooling a building as a function of time by evaluating $C (T (t)) .$ We have defined a new function, denoted $C \circ T,$ which is defined such that $(C \circ T) (t) = C (T (t))$ for all $t$ in the domain of $T .$ This new function is called a composite function. We note that since cost is a function of temperature and temperature is a function of time, it makes sense to define this new function $(C \circ T) (t) .$ It does not make sense to consider $(T \circ C) (t),$ because temperature is not a function of cost.

Definition

Consider the function $f$ with domain $A$ and range $B,$ and the function $g$ with domain $D$ and range $E .$ If $B$ is a subset of $D,$ then the composite function $(g \circ f) (x)$ is the function with domain $A$ such that

(g \circ f) (x) = g (f (x)) .

A composite function $g \circ f$ can be viewed in two steps. First, the function $f$ maps each input $x$ in the domain of $f$ to its output $f (x)$ in the range of $f .$ Second, since the range of $f$ is a subset of the domain of $g,$ the output $f (x)$ is an element in the domain of $g,$ and therefore it is mapped to an output $g (f (x))$ in the range of $g .$ In [link] , we see a visual image of a composite function.

An image with three items. The first item is a blue bubble that has two labels: “domain of f” and “domain of g of f”. This item contains the numbers 1, 2, and 3. The second item is two bubbles: an orange bubble labeled “domain of g” and a blue bubble that is completely contained within the orange bubble and is labeled “range of f”. The blue bubble contains the numbers 0 and 1, which are thus also contained within the larger orange bubble. The orange bubble contains two numbers not contained within the smaller blue bubble, which are 2 and 3. The third item is two bubbles: an orange bubble labeled “range of g” and a blue bubble that is completely contained within the orange bubble and is labeled “range of g of f”. The blue bubble contains the numbers 4 and 5, which are thus also contained within the larger orange bubble. The orange bubble contains one number not contained within the smaller blue bubble, which is the number 3. The first item points has a blue arrow with the label “f” that points to the blue bubble in the second item. The orange bubble in the second item has an orange arrow labeled “g” that points the orange bubble in the third item. The first item has a blue arrow labeled “g of f” which points to the blue bubble in the third item. There are three blue arrows pointing from numbers in the first item to the numbers contained in the blue bubble of the second item. The first blue arrow points from the 1 to the 0, the second blue arrow points from the 2 to the 1, and the third blue arrow points from the 3 to the 0. There are 4 orange arrows pointing from the numbers contained in the orange bubble in the second item, including those also contained in the blue bubble of the second item, to the numbers contained in the orange bubble of the third item, including the numbers in the blue bubble of the third item. The first orange arrow points from 2 to 3, the second orange arrow points from 3 to 5, the third orange arrow points from 0 to 4, and the fourth orange arrow points from 1 to 5. — For the composite function $g \circ f,$ we have $(g \circ f) (1) = 4, (g \circ f) (2) = 5,$ and $(g \circ f) (3) = 4 .$

Compositions of functions defined by formulas

Consider the functions $f (x) = x^{2} + 1$ and $g (x) = 1 / x .$

Find $(g \circ f) (x)$ and state its domain and range.
Evaluate $(g \circ f) (4), (g \circ f) (−1 / 2) .$
Find $(f \circ g) (x)$ and state its domain and range.
Evaluate $(f \circ g) (4), (f \circ g) (−1 / 2) .$

We can find the formula for $(g \circ f) (x)$ in two different ways. We could write
$(g \circ f) (x) = g (f (x)) = g (x^{2} + 1) = \frac{1}{x^{2} + 1} .$

Alternatively, we could write
$(g \circ f) (x) = g (f (x)) = \frac{1}{f (x)} = \frac{1}{x^{2} + 1} .$

Since $x^{2} + 1 \neq 0$ for all real numbers $x,$ the domain of $(g \circ f) (x)$ is the set of all real numbers. Since $0 < 1 / (x^{2} + 1) \leq 1,$ the range is, at most, the interval $(0, 1] .$ To show that the range is this entire interval, we let $y = 1 / (x^{2} + 1)$ and solve this equation for $x$ to show that for all $y$ in the interval $(0, 1],$ there exists a real number $x$ such that $y = 1 / (x^{2} + 1) .$ Solving this equation for $x,$ we see that $x^{2} + 1 = 1 / y,$ which implies that
$x = \pm \sqrt{\frac{1}{y} - 1} .$

If $y$ is in the interval $(0, 1],$ the expression under the radical is nonnegative, and therefore there exists a real number $x$ such that $1 / (x^{2} + 1) = y .$ We conclude that the range of $g \circ f$ is the interval $(0, 1] .$
$(g \circ f) (4) = g (f (4)) = g (4^{2} + 1) = g (17) = \frac{1}{17}$
$(g \circ f) (- \frac{1}{2}) = g (f (- \frac{1}{2})) = g ({(- \frac{1}{2})}^{2} + 1) = g (\frac{5}{4}) = \frac{4}{5}$
We can find a formula for $(f \circ g) (x)$ in two ways. First, we could write
$(f \circ g) (x) = f (g (x)) = f (\frac{1}{x}) = {(\frac{1}{x})}^{2} + 1 .$

Alternatively, we could write
$(f \circ g) (x) = f (g (x)) = {(g (x))}^{2} + 1 = {(\frac{1}{x})}^{2} + 1 .$

The domain of $f \circ g$ is the set of all real numbers $x$ such that $x \neq 0 .$ To find the range of $f,$ we need to find all values $y$ for which there exists a real number $x \neq 0$ such that
${(\frac{1}{x})}^{2} + 1 = y .$

Solving this equation for $x,$ we see that we need $x$ to satisfy
${(\frac{1}{x})}^{2} = y - 1,$

which simplifies to
$\frac{1}{x} = \pm \sqrt{y - 1} .$

Finally, we obtain
$x = \pm \frac{1}{\sqrt{y - 1}} .$

Since $1 / \sqrt{y - 1}$ is a real number if and only if $y > 1,$ the range of $f$ is the set ${y | y \geq 1} .$
$(f \circ g) (4) = f (g (4)) = f (\frac{1}{4}) = {(\frac{1}{4})}^{2} + 1 = \frac{17}{16}$
$(f \circ g) (- \frac{1}{2}) = f (g (- \frac{1}{2})) = f (−2) = {(−2)}^{2} + 1 = 5$

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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