Question

34. Consider f: R. → [4.-) given by f(x) = x2 + 4. Show that f is invertible with the inverse f- off given by f-(y) = y-4, where R. is the set of all non- negative real numbers. ​

Answer 1

Given :-

${f}^{ - 1} (y) = \sqrt{y - 4}$

To Find The Solution :-

f.R + => [4,oo) is given as f(x) = x²+ 4 .

One-One :-

Let's f(x) = f(y)

${x}^{2} + 4 = {y}^{2} + 4$

${x}^{2} = {y}^{2}$

$x = y$

[as x = y € R ±]

•f is a One-One function .

Onto :

For y € [4,oo), Let y = x²+4

${x}^{2} = y - 4≥0$

$x = \sqrt{y - 4} ≥0$

[As y ≥ 4]

Therefore, for any y € R, there exists :-

$x = \sqrt{y - 4} €R \:$

Such That.

$f(x) = f( \sqrt{y - 4} ) = ( \sqrt{y - 4) {}^{2} } + 4 =y - 4 + 4 = y$

∴ f is Onto.

Thus,f is One-One and onto and therefore f-¹ exists .

Let us define g: [4,oo) => R± by,

$g(y) = \sqrt{y - 4}$

$now.gof(x) = g(f(x)) = g( {x}^{2} + 4) = \sqrt{(x2 + 4)} - 4 = \sqrt{ {x}^{2} } = x$

$and.fog (y) = f(g(y)) = f( \sqrt{(y - 4)} = ( \sqrt{(y - 4)} {}^{2} + 4 = (y - 4) + 4 = y$

•gof = fog = IR±

Hence,f is invertible and the inverse of f is given by

$f { }^{ - 1} (y) = g(y) = \sqrt{y - 4.}$

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