Question

Given the function }" alt="h:x=px-\frac{5}{2}" align="absmiddle" class="latex-formula"> and the inverse function $h^{-1} :x=q+2x$, where p and q are constants, find a) the value of p and q c)$h^{-1} h(-3)$

Answer 1

Answer:

$p = \frac{1}{2}$

$q = 5$

$h^{-1}(h(3)) = 3$

Step-by-step explanation:

Given

$h(x) = px - \frac{5}{2}$

$h^{-1}(x) = q + 2x$

Solving for p and q

Replace h(x) with y in $h(x) = px - \frac{5}{2}$

$y = px - \frac{5}{2}$

Swap the position of y and d

$x = py - \frac{5}{2}$

Make y the subject of formula

$py = x + \frac{5}{2}$

Divide through by p

$y = \frac{x}{p} + \frac{5}{2p}$

Now, we've solved for the inverse of h(x);

Replace y with $h^{-1}(x)$

$h^{-1}(x) = \frac{x}{p} + \frac{5}{2p}$

Compare this with $h^{-1}(x) = q + 2x$

We have that

$\frac{x}{p} + \frac{5}{2p} = q + 2x$

By direct comparison

$\frac{x}{p} = 2x$ --- Equation 1

$\frac{5}{2p} = q$  --- Equation 2

Solving equation 1

$\frac{x}{p} = 2x$

Divide both sides by x

$\frac{1}{p} = 2$

Take inverse of both sides

$p = \frac{1}{2}$

Substitute $p = \frac{1}{2}$ in equation 2

$\frac{5}{2 * \frac{1}{2}} = q$

$\frac{5}{1} = q$

$5 = q$

$q = 5$

Hence, the values of p and q are:$p = \frac{1}{2}$;  $q = 5$

Solving for $h^{-1}(h(3))$

First, we'll solve for h(3) using $h(x) = px - \frac{5}{2}$

Substitute $p = \frac{1}{2}$;  and $x = 3$

$h(3) = \frac{1}{2} * 3 - \frac{5}{2}$  

$h(3) = \frac{3}{2} - \frac{5}{2}$

$h(3) = \frac{3 - 5}{2}$

$h(3) = \frac{-2}{2}$

$h(3) = -1$

So; $h^{-1}(h(3))$ becomes

$h^{-1}(-1)$

Solving for $h^{-1}(-1)$ using $h^{-1}(x) = q + 2x$

Substitute $q = 5$ and $x = -1$

$h^{-1}(x) = q + 2x$ becomes

$h^{-1}(-1) = 5 + 2 * -1$

$h^{-1}(-1) = 5 - 2$

$h^{-1}(-1) = 3$

Hence;

$h^{-1}(h(3)) = 3$