Question

Determine whether each function has an inverse function. If it does, find the inverse function and state any restrictions on its domain.
1. $f(x) = |x-6|$

2. $f(x) = \sqrt[]{6 -x^{2} }$

3. $h(x) = \frac{x+4}{3x-5}$

Answer 1

QUESTION 1

The given function is;

$f(x) = |x - 6|$

This is an absolute value function.

For this function to have an inverse it must be a one-to-one function.

This absolute value function is not one-to-one
because

$f( 5)=1$
and

$f(7) = 1$

Since this function has more than one x-value mapping onto one y-value, it is not one-to-one and cannot have an inverse.

You can see from the graph that this function cannot pass the horizontal line test.

QUESTION 2

The given function is

$f(x) = \sqrt{6 - {x}^{2} }$

Let

$y= \sqrt{6 - {x}^{2} }$

This implies that,

${y}^{2} + {x}^{2} = 6$

We see clearly that, this function is a circle that is centered at the origin.

This means that,

$f(x) = \sqrt{6 - {x}^{2} }$

is a semicircle.

This function will not pass the horizontal line test and hence does not have an inverse.

QUESTION 3

The given function is

$h(x) = \frac{x+4}{3x-5}$

If we put

$h(a) = h(b)$

We obtain,

$\frac{a+4}{3a-5} = \frac{b+4}{3b-5}$

$(a + 4)(3b - 5) = (b + 4)(3a - 5)$

$3ab - 5a + 12b - 20 = 3ab - 5b + 12a - 20$

$- 17a = - 17b$

$a = b$

This shows that h(x) has an inverse because it is one-to-one.

Let

$y=\frac{x+4}{3x-5}$

We interchange x and y to get,

$x=\frac{y+4}{3y-5}$

$x(3y - 5) = y + 4$

$3xy - 5x = y + 4$

Group like terms to get,

$3xy - y = 5x + 4$

Factor to get,

$y(3x - 1) = 5x + 4$
Solve for y,

$y = \frac{5x + 4}{3x - 1}$

Hence

${f}^{ - 1}(x) = \frac{5x + 4}{3x - 1}$

where

$x \ne \frac{1}{3}$