- From: Kelly Biddington
- Date: 22 April 1999
- Subject: Inverse Functions
The function f is given by f:x --> ln(4 - 2x), x<2 - Find an expression for f^-1(x)
- State the range of f^-1
- Sketch the graph of y = f^-1(x)
When it comes to e and ln, I become all quivery at the knees!!! |

Just as addition and subtraction are inverse processes (each 'undoes' the other),

and squaring and square rooting are inverse proceses,

e^{x} and ln *x* are inverse processes.

To solve your problem, let *y* = f(*x*) then make *x* the subject:

This function of y is the form of the inverse of the original function of x.

Write the inverse function as

The range of the inverse function f^{-1}(*x*) is the domain of f(*x*),
namely *x* < 2.

To make a successful sketch of the inverse function, you should work out

the *y*-intercept and get the general shape right:

Convince yourself that this can be built up from the basic graph of **y = e ^{x}**
as follows:

- Stretch vertically, scale factor 0.5 to give
**y = 0.5e**^{x} - Reflect in x-axis to give
**y = - 0.5e**^{x} - Translate vertically by 2 units upwards to give
**y = 2 - 0.5e**^{x}

REMEMBER that a useful way to get the graph of an inverse function is to reflect the
graph of the original function in the line y=x. We have not done that here for two reasons.

Firstly, the graph of f(x) is itself not so straightforward to sketch (it needs a series of
transformations of the basic graph y = ln(x)). And secondly, in this particular question, the graph
of f(x) and its inverse overlap quite considerably. We suggest you investigate them using a
graphics calculator or computer graph plotter.

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