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amyl520

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Calculus I - Derivatives of Inverse Function

For the function y = f(x) = (x^2) - 3x - 1, x greater than or equal to 1.5, find (df^-1)/dy when y=-2.

(f^-1)'(-2) = ?

Bob

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Registered: 2010-06-20

Posts: 10,640

Re: Calculus I - Derivatives of Inverse Function

hi amyl520

Welcome to the forum.

I'll explain the background to this first.

If you plot the graph of f against x you get a typical quadratic.

The inverse wouldn't be a function because for every value of y you get more than 1 value of x.  Functions have to have a single value.

But the question limits us to x ≥ 1.5.  If you plot the graph

https://www.mathsisfun.com/data/function-grapher.php

you'll see that x = 1.5 is the minimum point so the limit means we have a single valued inverse.

When you differentiate you are finding the gradient function, in the case dy/dx = 2x - 3

If the graph is reflected in the line y=x, the x and y axes swap over and so we have x as a function of y.  The gradient function for this graph is just dx/dy so all you need to do is invert the gradient function, ie. 1 / (2x-3)

I tried this and it didn't come out to be a simple integer:

But you're asked for the value when y = -2, so you have to work out what x is, when y = -2

( the minus case from the quadratic formula is out of  the domain x ≥ 1.5)

So you need to evaluate:

Hope that helps,

Bob

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Children are not defined by school ...........The Fonz
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