QUESTION 1
The given function is
[tex]f(x) = \frac{2x}{3} - 17[/tex]
To find the inverse function, we let
[tex]y = f(x)[/tex]
This implies that,
[tex]y= \frac{2x}{3} - 17[/tex]
This will give us,
[tex]y + 17= \frac{2x}{3} [/tex]
We multiply through by 3 to obtain;
[tex]3(y + 17) = 2x [/tex]
We now interchange x and y to obtain,
[tex]3x = 2y - 51[/tex]
We make y the subject to obtain,
[tex]3(x + 17) = 2y[/tex]
[tex]y = \frac{3(x + 17)}{2} [/tex]
This implies that,
[tex] {f}^{ - 1} (x) = \frac{3(x + 17)}{2} [/tex]
Therefore,
[tex]f(x) = \frac{2x}{3} - 17 \rightarrow \: {f}^{ - 1} (x) = \frac{3(x + 17)}{2} [/tex]
QUESTION 2
The given function is
[tex]f(x) = x - 10[/tex]
To find the inverse function we let
[tex]y = x - 10[/tex]
We then interchange x and y to obtain,
[tex]x = y - 10[/tex]
We solve for y to obtain,
[tex]y = x + 10[/tex]
Therefore the inverse function is
[tex] {f}^{ - 1} (x) = x + 10[/tex]
Hence,
[tex] f(x) = x - 10 \rightarrow \: {f}^{ - 1} (x) = x + 10[/tex]
QUESTION 3.
The given function is
[tex]f(x) = \sqrt[3]{2x} [/tex]
We want to find the inverse so we let
[tex]y=\sqrt[3]{2x} [/tex]
We now interchange x and y to obtain,
[tex]x=\sqrt[3]{2y} [/tex]
We now make y the subject, by first taking the cube of both sides of the equation.
[tex] {x}^{3} = 2y[/tex]
Divide through by 2 to get,
[tex] \frac{ {x}^{3} }{2} = y[/tex]
Or
[tex] y = \frac{ {x}^{3} }{2}[/tex]
This implies that,
[tex] {f}^{ - 1}(x) = \frac{ {x}^{3} }{2} [/tex]
Therefore
[tex] f(x) = \sqrt[3]{2x} \rightarrow \: {f}^{ - 1}(x) = \frac{ {x}^{3} }{2} [/tex]
QUESTION 4
The given function is
[tex]f(x) = \frac{x}{5} [/tex]
We let
[tex]y = \frac{x}{5} [/tex]
Interchange x and y to get,
[tex]x = \frac{y}{5} [/tex]
Make y the subject to get,
[tex]y = 5x[/tex]
This implies that,
[tex] {f}^{ - 1} (x)= 5x[/tex]
[tex] f(x) = \frac{x}{5} \: \rightarrow \: {f}^{ - 1} (x)= 5x[/tex]
