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# Calculus/Pre- calculus

Question
Find the inverse function of f(x)=5x/2x-3 and find the inverse function of f(x)=(2x+1)^5/2 -3

It looks like 5x/2x - 3 is really 5x/(2x) - 3, which is 5/2 - 3 = -1/2.
This has no inverse function.

Is this suppose to be f(x) = (2x+1)^(5/2) - 3?
If so, since 5/2 = 2.5, make it f(x) = (2x+1)^2.5 - 3.

To get the inverse, repalce f(x) with y and solve for x.
That is, take y = (2x+1)^2.5 - 3 and start by adding 3 to both sides.
This gives y + 3 = (2x+1)^2.5.  Since 2.5 * 0.4 = 1, take both sides to the 0.4.
This gives (y+3)^0.4 = 2x + 1.  To get x, subtract 1 and then divide by 2.
This gives x = [(y+3)^0.4 - 1]/2.

To make it the inverse function, change the labelling of x and y, giving
y = [(x+3)^0.4 - 1]/2.  Then replace the y with a g(x), so g(x) = [(x+3)^0.4 - 1]/2.

We now have g(x) = f^-1(x) or f(x) = g^-1(x).
The requirements to keep this a real function is 2x+1>=0, so x>-1/2.
When x=-1/2, y=3, so that is the minimum for y.

Calculus

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