# What is the inverse function of h(x)= 3-(x+4)/(x-7) and how do you evaluate h^-1(9)?

Jun 11, 2015

${h}^{- 1} \left(y\right) = \frac{7 y - 25}{y - 2}$ where $y \ne 2$.
${h}^{- 1} \left(9\right) = \frac{38}{7}$

#### Explanation:

$h \left(x\right) = 3 - \frac{x + 4}{x - 7}$ having $x \ne 7$
I find the common denominator and sum all together:
$h \left(x\right) = \frac{3 x - 21}{x - 7} - \frac{x + 4}{x - 7}$
$h \left(x\right) = \frac{2 x - 25}{x - 7}$
Now I simplify making the division of the 2 polynomials and I obtain:
$\text{quotient"=2, "reminder} = - 11$
So I can write the function as:
$h \left(x\right) = 2 \frac{x - 7}{x - 7} - \frac{11}{x - 7}$
$h \left(x\right) = 2 - \frac{11}{x - 7}$.

Now, the question is how to find the inverse function? Firstly, I try to isolate x:
$\left(x - 7\right) \left(h \left(x\right) - 2\right) = - 11$
$\left(x - 7\right) = - \frac{11}{h \left(x\right) - 2}$
$x = - \frac{11}{h \left(x\right) - 2} + 7$
Therefore we rewrite better the function:
${h}^{- 1} \left(y\right) = - \frac{11}{y - 2} + 7 = \frac{7 y - 14 - 11}{y - 2} = \frac{7 y - 25}{y - 2}$.
So we can state that:
${h}^{- 1} \left(y\right) = \frac{7 y - 25}{y - 2}$ where $y \ne 2$.

If we want to find ${h}^{- 1} \left(9\right)$:
${h}^{- 1} \left(9\right) = \frac{7 \cdot 9 - 25}{9 - 2} = \frac{63 - 25}{7} = \frac{38}{7}$

Jun 11, 2015

The inverse function is ${h}^{- 1} \left(x\right) = \frac{7 x - 25}{x - 2}$.
${h}^{- 1} \left(x\right) = \frac{38}{7}$

#### Explanation:

Since the original function is not that complex, you can determine its inverse function faster by solving the function for $x$ and switching the result for $h \left(x\right)$.

$h \left(x\right) = 3 - \frac{x + 4}{x - 7}$

$h \left(x\right) = \frac{3 \left(x - 7\right) - \left(x + 4\right)}{x - 7} = \frac{3 x - 21 - x - 4}{x - 7}$

$h \left(x\right) = \frac{2 x - 25}{x - 7}$

Solve this form of the equation for $x$ to get

${h}^{x} \cdot \left(x - 7\right) = 2 x - 25$

$x \cdot h \left(x\right) - 7 \cdot h \left(x\right) \cdot 7 = 2 x - 25$

$x \cdot h \left(x\right) - 2 x = 7 \cdot h \left(x\right) - 25$

$x \left(h \left(x\right) - 2\right) = 7 \cdot h \left(x\right) - 25$

$x = \frac{7 \cdot h \left(x\right) - 25}{h \left(x\right) - 2}$

Once you isolate $x$, simply switch $h \left(x\right)$ for $x$ to get the inverse function

${h}^{- 1} \left(x\right) = \frac{7 x - 25}{x - 2}$

To evaluate ${h}^{- 1} \left(9\right)$, simply substitute $x$ with $9$ to get

${h}^{- 1} \left(9\right) = \frac{7 \cdot 9 - 25}{9 - 2} = \frac{38}{7}$