Need help finding an inverse function

• Sep 2nd 2011, 08:14 AM
Algebrah
Need help finding an inverse function
I have a function that describes a ball being thrown at an initial velocity of 192 ft/s, given by:

f(t) = 192t - 16t^2

I need to find an inverse function that yields the time 't' when the ball is at a height 'h' as the object travels upward:

t = 6 - [sqrt(576 - h) / 4]

Also, I need to find an inverse function that yields the time 't' when the ball is at height 'h' as the object travels downward:

t = 6 + [sqrt(576 - h) / 4]

The problem is that I can't figure out how to find these inverse functions from f(t) = 192t - 16t^2. I tried factoring, but didn't have any luck. Also, since this is a second order function, it's not one-to-one, which means that it shouldn't have an inverse, right (since it will fail the horizontal line test)? The two answers that were given above were what was listed, but I can't figure out how to get to the answer. (Worried)

Help would be greatly appreciated! Thanks!
• Sep 2nd 2011, 09:18 AM
Siron
Re: Need help finding an inverse function
When the ball is at height $\displaystyle h$ that means $\displaystyle f(t)=h$ so the equation becomes:
$\displaystyle h=192t-16t^2$
$\displaystyle \Leftrightarrow 16t^2-192t+h=0$
Calculating the discriminant gives:
$\displaystyle D=(-192)^2-4\cdot 16 \cdot h = 36 864 - 64h=64(576-h)$

So the two solutions are ...
• Sep 2nd 2011, 12:57 PM
HallsofIvy
Re: Need help finding an inverse function
Siron is referring to the "quadratic formula".

Solutions to $\displaystyle ax^2+ bx+ c= 0$ are given by
$\displaystyle \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$.

Your function, $\displaystyle h= 192t- 16t^2$ is the same as $\displaystyle 16t^2- 192t+ h= 0$ so a= 16, b= -192, c= h.
• Sep 3rd 2011, 03:57 PM
Re: Need help finding an inverse function
Or you can use the "completing the square" method.

First make 16t^2-192t the subject:
16t^2-192t = -h

Factorise to make it easier:
16(t^2-12t) = -h

Completing the square gives:
16((t-6)^2-36) = -h

You should be able to take it from here...

It's the same as HallofIvy's only different approach...