[SOLVED] Solving for inverse functions: f(x)=x^2-5x-6

Jun 2015
2
0
Toronto
I'm trying to find the inverse function of f(x)=x^2-5x-6

All help is appreciated!

Edit: Figured it out!
 
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HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
In case others are interested- the first thing I would do is "complete the square". \(\displaystyle y= x^2- 5x- 6= x^3- 5x+ \frac{25}{4}- \frac{25}{4}- 6\)\(\displaystyle = \left(x- \frac{5}{2}\right)^2- \frac{49}{4}\). That tells me that the vertex of this parabolic graph is at \(\displaystyle \left(\frac{5}{2}, \frac{49}{4}\right)\). So, while the entire function does NOT have a true "inverse" I can restrict the domain to get two functions that have inverses. From \(\displaystyle y= \left(x- \frac{5}{2}\right)^2- \frac{49}{4}\), \(\displaystyle \left(x- \frac{5}{2}\right)^2= y+ \frac{49}{4}\) and then \(\displaystyle x= \frac{5}{4}\pm\sqrt{y+ \frac{49}{4}}\). The part with \(\displaystyle x< \frac{5}{4}\) is given with the "-" and that with \(\displaystyle x> \frac{5}{4}\) is given with the "+".

Finally, swapping "x" and "y", the two inverses are given by \(\displaystyle y= \frac{5}{4}+ \sqrt{x+ \frac{49}{4}\) and \(\displaystyle y= \frac{5}{4}- \sqrt{x+ \frac{49}{4}\).
 
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