Inverse Function

This post will focus in solving inverse function. Another easy topic that we need to master in order for us solve harder problem that requires our skill with inverses. The only rule we follow  to solve for the inverse of a function is to interchange x and y and solve for y.

Worked Problem 1:

Find the inverse $f(x)=3x+2$

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We know that let $f(x)=y$

$y=3x+2$

Interchange x and y:

$x=3y+2$

Solve for y:

$x-2=3y$ $3y=x-2$

$y=\displaystyle\frac{x-2}{3}$ [/toggle]

Worked Problem 2:

What is the inverse of $y=5^x-2$

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Interchange x and y:

$x=5^y-2$ $5^y=x+2$

Take note that y is the exponent of base 5.

According to rule of logarithm, $\log x^n=n\log x$ and this is our only hope to bring y at the bottom.

We take the logs of both sides to keep the equality.

$\log 5^y=\log (x+2)$ $y\log 5=\log (x+2)$ $y=\displaystyle\frac{\log (x+2)}{\log 5}$

According to change of base formula of logarithm,

$\log_xy=\displaystyle\frac{\log_by}{log_bx}$

$y=log_5(x+2)$ [/toggle]

Worked Problem 3:

Find the inverse function of $y=x^2-2x+3$

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Interchange x and y:

$x=y^2-2y+3$ $y^2-2y+3-x=0$

Treat x as constant. Note that the equation now is quadratic in y. We can use quadratic formula to solve for y with the following values of a,b, and c.

$a=1, b=-2, c=3-x$ $y=\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$y=\displaystyle\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(3-x)}}{2(1)}$    by substitution

$y=\displaystyle\frac{2\pm\sqrt{(4-12+4x)}}{2}$ $y=\displaystyle\frac{2\pm\sqrt{4x-8}}{2}$ $y=\displaystyle\frac{2\pm\sqrt{4(x-2)}}{2}$ $y=\displaystyle\frac{2\pm 2\sqrt{x-2}}{2}$

$y=1\pm\sqrt{x-2}$ [/toggle]

But wait! There’s more. I will leave you with a problem to challenge yourself.

John Hyperactive is having fun with inverses. He get the inverse of quadratic function $f(x)=x^2-9$. He solved the inverse of the function and after getting it. He solved further the inverse of the inverse function. He repeated the process 2014th times. What equation he had right now?