QUESTION 1
The given function is;
[tex]f(x) = |x - 6| [/tex]
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
[tex]f( 5)=1[/tex]
and
[tex]f(7) = 1[/tex]
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
[tex]f(x) = \sqrt{6 - {x}^{2} } [/tex]
Let
[tex]y= \sqrt{6 - {x}^{2} } [/tex]
This implies that,
[tex] {y}^{2} + {x}^{2} = 6[/tex]
We see clearly that, this function is a circle that is centered at the origin.
This means that,
[tex]f(x) = \sqrt{6 - {x}^{2} } [/tex]
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
[tex]h(x) = \frac{x+4}{3x-5}[/tex]
If we put
[tex]h(a) = h(b)[/tex]
We obtain,
[tex]\frac{a+4}{3a-5} = \frac{b+4}{3b-5}[/tex]
[tex](a + 4)(3b - 5) = (b + 4)(3a - 5)[/tex]
[tex]3ab - 5a + 12b - 20 = 3ab - 5b + 12a - 20[/tex]
[tex] - 17a = - 17b[/tex]
[tex]a = b[/tex]
This shows that h(x) has an inverse because it is one-to-one.
Let
[tex]y=\frac{x+4}{3x-5}[/tex]
We interchange x and y to get,
[tex]x=\frac{y+4}{3y-5}[/tex]
[tex]x(3y - 5) = y + 4[/tex]
[tex]3xy - 5x = y + 4[/tex]
Group like terms to get,
[tex]3xy - y = 5x + 4[/tex]
Factor to get,
[tex]y(3x - 1) = 5x + 4[/tex]
Solve for y,
[tex]y = \frac{5x + 4}{3x - 1} [/tex]
Hence
[tex] {f}^{ - 1}(x) = \frac{5x + 4}{3x - 1} [/tex]
where
[tex]x \ne \frac{1}{3} [/tex]
