Question

Part 1. Using the two functions listed below, insert numbers in place of the letters a, b, c, and d so that f(x) and g(x) are inverses. f(x)= x+a b g(x)=cx−d

Answer 1

Answer:

c = 1 and a,d can take any value such that a = d

Step-by-step explanation:

Function is a relative in which every element of domain has a unique image in the co-domain.

An inverse of a function exists if it is one-one and onto.

A function is onto if every element of the co-domain has a preimage in domain.

A function is said to be one to one if elements of the domain have different images in the co-domain.

$y=f(x)=x+a\\y=x+a\\x=y-a\\f^{-1}(x)=x-a$

On comparing $f^{-1}(x)=x-a$ and $g(x)=cx-d$, we get c = 1 and a = d.

So, c = 1 and a,d can take any value such that a = d

Answer 2

Answer:

The options are endless.

You do have the following requirements to the letters asked about:

c=b  (neither of these can be 0)

d=a

So here is one example:

Choose c=b=4 and d=a=9.

The functions will be:

$f(x)=\frac{x+9}{4}$ and

$g(x)=4x-9$

Step-by-step explanation:

I'm going to assume my comment above.

In order for f and g to be inverse f(g(x)) has to give you x and g(f(x)) also has to give you x.

$f(g(x))$

$f(cx-d)$

$\frac{(cx-d)+a}{b}$

$\frac{cx-d+a}{b}$

$\frac{cx+(-d+a)}{b}$

$\frac{cx}{b}+\frac{-d+a}{b}$

We want this to equal x.

This means we need the following to happen:

$\frac{c}{b}=1 \text{ and } \frac{-d+a}{b}=0$

The first one happens when $c=b$ assuming b isn't 0 since you can't divide by 0.

The second one happens when the top is 0 because that is the only way a fraction will output you 0 is if the numerator is 0.

So we $d=a$ as well.

So this is what we need for f and g so that they are inverse:

$f(x)=\frac{x+d}{c}$ since $d=a$ and $c=b$

$g(x)=cx-d$

Let's try to see if we get x when doing both f(g(x)) and g(f(x)).

$f(g(x))$

$f(cx-d)$

$\frac{(cx-d)+d}{c}$

$\frac{cx-d+d}{c}$

$\frac{cx}{c}$

$x$

We have confirmation that f(g(x))=x.

$g(f(x))$

$g(\frac{x+d}{c})$

$c(\frac{x+d}{c})-d$

$x+d-d$

$x$

So $f(x)=\frac{x+d}{c}$ and $g(x)=cx-d$ are inverses for all choices c and d except when c is 0 since you are not allowed to divide by 0.

Your question says to insert numbers for c (=b) and d (=a).

So an example:

Let b=c=4 and d=a=9.

Let's try it:

These will be our functions with those choices for the constant variables:

$f(x)=\frac{x+9}{4}$ and $g(x)=4x-9$

$f(g(x))$

$f(4x-9)$

$\frac{(4x-9)+9}{4}$

$\frac{4x}{4}$

$x$

We have f(g(x))=x.  Check mark, there.

$g(f(x))$

$g(\frac{x+9}{4})$

$4(\frac{x+9}{4})-9$

$x+9-9$

$x$