Inverse of a Function

The functions

f (x) = 2 x - 3

and

f^{- 1} (x) = \frac{x + 3}{2}

will be shown to undo each other. To do so, it needs to be proven that

f (f^{- 1} (x)) = x

and that

f^{- 1} (f (x)) = x .

To start, the first equality will be proven. First, the definition of

f

will be used.

f (f^{- 1} (x)) = ? x ⇕ 2 f^{- 1} (x) - 3 = ? x

Now, in the above equation,

\frac{x + 3}{2}

will be substituted for

f^{- 1} (x) .

2 f^{- 1} (x) - 3 = ? x

Substitute

$f^{- 1} (x) = \frac{x + 3}{2}$

2 (\frac{x + 3}{2}) - 3 = ? x

▼

Simplify left-hand side

DenomMultFracToNumber

$2 \cdot \frac{a}{2} = a$

(x + 3) - 3 = ? x

RemovePar

Remove parentheses

x + 3 - 3 = ? x

SubTerm

Subtract term

x = x ✓

A similar procedure can be performed to show that

f^{- 1} (f (x)) = x .

	Definition of First Function	Substitute Second Function	Simplify
$f (f^{- 1} (x)) = ? x$	$2 f^{- 1} (x) - 3 = ? x$	$2 (\frac{x + 3}{2}) - 3 = ? x$	$x = x ✓$
$f^{- 1} (f (x)) = ? x$	$\frac{f ( x ) + 3}{2} = ? x$	$\frac{2 x - 3 + 3}{2} = ? x$	$x = x ✓$

Therefore, $f$ and $f^{- 1}$ undo each other. The graphs of these functions are each other's reflection across the line $y = x .$ This means that the points on the graph of $f^{- 1}$ are the reversed points on the graph of $f .$

$Graph of the linear function f(x)=2*x-3 with a point (a,b) on it and the graph of its inverse function f^{-1}(x)=x/2+3/2 with a point (b,a) on it are depicted. The line of symmetry x=y for these functions is also shown.$

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Inverse of a Function

Why

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