by Guest » Wed Dec 01, 2021 8:21 pm
That's pretty straight forward. It's hard to see why you would call that "challenging" since you don't show what you tried.
We can write [tex]f(x)= y= \frac{5x^2+ 8}{x+ 7}[/tex] and get the inverse function by swapping x and y: [tex]x= \frac{5y^2+ 8}{y+ 7}[/tex] and get to the form [tex]y= f^{-1}(x)[/tex] by solving for y.
[tex]x(y+ 7)= xy+ 7x= 5y^2+ 8[/tex]
[tex]5y^2+ xy+ 7x- 8= 0[/tex]
By the quadratic formula
[tex]y= \frac{-x\pm\sqrt{x^2- 140x+ 160}}{10}[/tex]
Now the problem is that "[tex]\pm[/tex]". The original function may not HAVE an inverse. In that case we would need to separate f into two functions, each one of which would have an inverse.