Inverse and composite functions

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Matthew Mears

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Matthew Mears 6 years, 5 months ago

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Give all of your answers in terms of $x$ .

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			Name	Type	Generated Value

			a	list	[ 5, 2, 5, 2, 4 ]
			b	list	[ 8, 9, 6, 9, 5 ]

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xxxxxxxxxx
 
repeat(random(2..5),5)

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Generated value: `list`

[ 5, 2, 5, 2, 4 ]

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Part b) Gap-fill
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Part e) Gap-fill
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Part f) Choose one from a list

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Find $f^{-1}(x)$ when $\simplify{f(x)={a[0]}x+{a[1]} }$ .

$\displaystyle f^{-1}(x)=$ Unnamed gap

Advice

a)

$f^{-1}(x)$ is the function with the property that $f^{-1}(f(x)) = x$ .

To find this, we first set $x=f(y)$ and rearrange to find $y$ in terms of $x$ , i.e. $y = f^{-1}(x)$ .

$\begin{align} f(y)=\simplify{{a[0]}y+{a[1]}}&=x\\ \simplify{{a[0]}y}&=x-\var{a[1]}\\[1em] y&=\simplify[]{(x-{a[1]})/{a[0]}}\\[1em] f^{-1}(x)&=\simplify{(x-{a[1]})/{a[0]}}\text{.}\\ \end{align}$

b)

We use the same method as part a) to find $g^{-1}(x)$ :

$\begin{align} g(y)=\simplify{{a[3]}y-{a[2]}}&=x\\ \simplify{{a[3]}y}&=x+\var{a[2]}\\[1em] y&=\simplify[]{(x+{a[2]})/{a[3]}}\\[1em] g^{-1}(x)&=\simplify{(x+{a[2]})/{a[3]}}\text{.}\\ \end{align}$

c)

$(g^{-1} \circ f^{-1})(x)$ is the function which first applies $f^{-1}(x)$ and then applies $g^{-1}$ to the result of that.

We use the previous answers: $f^{-1}(x)=\simplify{(x-{a[1]})/{a[0]}}$ and $g^{-1}(x)=\simplify{(x+{a[2]})/{a[3]}}$ to find the definition of $(g^{-1} \circ f^{-1})(x)$ by substituting $f^{-1}(x)$ everywhere $x$ occurs in the definition of $g^{-1}(x)$ .

$\begin{align} (g^{-1}\circ f^{-1}) (x)&=g^{-1}(f^{-1}(x))\\[1em] &=g^{-1} \left( \simplify[]{(x-{a[1]})/{a[0]}} \right) \\[1em] &=\frac{\left(\simplify[]{(x-{a[1]})/{a[0]}}\right)+\simplify[]{{a[2]}}}{\var{a[3]}\text{.}} \end{align}$

d)

$(f \circ g)(x)$ is the function which first applies $g(x)$ and then applies $f$ to the result of that.

We find the definition of $(f \circ g)(x)$ by substituting $g(x)$ everywhere that $x$ occurs in the definition of $f(x)$ .

$\begin{align} (f\circ g)(x)&=f(g(x))\\ &=f(\simplify{{a[3]}x-{a[2]}})\\ &=\simplify{{a[0]}({a[3]}x-{a[2]})+{a[1]}} \end{align}$

e)

Now that we have the definition of $(f \circ g)(x)$ , we can find its inverse by using the same method as in parts a) and b).

$\begin{align} (f \circ g)(y) &= x \\ \simplify{{a[0]}({a[3]}y-{a[2]})+{a[1]}}&=x\\ \simplify{{a[0]}({a[3]}y-{a[2]})}&=x-\var{a[1]}\\[1em] \simplify{{a[3]}y-{a[2]}}&=\frac{(x-\var{a[1]})}{\var{a[0]}}\\[1em] \simplify{{a[3]}y}&=\left( \frac{(x-\var{a[1]})}{\var{a[0]}} \right) +\var{a[2]}\\[1em] y&=\frac{\left(\frac{(x-\var{a[1]})}{\var{a[0]}}\right)+\var{a[2]})}{\var{a[3]}}\\[1em] (f\circ g)^{-1}(x)&=\frac{\left(\frac{(x-\var{a[1]})}{\var{a[0]}}\right)+\var{a[2]}}{\var{a[3]}}\\[1em] \end{align}$

We can see that in this case $(f\circ g)^{-1}(x) = (g^{-1}\circ f^{-1}) (x)$ .

f)

So long as the inverses of $f$ and $g$ exist and they can be composed, it is always true that

$(f \circ g)^{-1}(x) \equiv (g^{-1} \circ f^{-1}) (x)\text{.}$

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Matthew Mears 6 years, 5 months ago

Error in variable testing condition

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Generated value: `list`

Parts

Gap-fill

a)

b)

c)

d)

e)

f)

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Inverse and composite functions

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Matthew Mears 6 years, 5 months ago

Error in variable testing condition

Error

Warning

Generated value: list

Parts

Gap-fill

a)

b)

c)

d)

e)

f)

Generated value: `list`