Answered: Z is the set of integers, R is the set…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Z is the set of integers, R is the set of real numbers.

4) Find the inverse of functions
a) f: [0,∞) -> [0,∞), f(x) = x² + 12;
b) f: R - {-5/2} -> R - {3/2}, f(x) = 3x/(2x + 5);
c) f: R⁺ -> (0,1), f(x) = 1/(x + 1);

Expert Solution

Step 1

(a)

Consider the function, $f : [0, \infty) \to [0, \infty)$ defined by $f (x) = x^{2} + 12$

For $x \in [0, \infty)$ , $f (x) \in [12, \infty]$ .

That is, range set is not equal to the co-domain.

So, the function is not onto, and therefore not invertible.

But, by restricting the the co-domain $[0, \infty)$ to $[12, \infty]$ , $f (x)$ becomes both one-one and onto and hence invertible.

Suppose $y = f (x)$ , then, $f^{- 1} = \{x | f (x) = y\}$

That is,

$\begin{array}{rcl} y & = & x^{2} + 12 \\ x^{2} & = & y - 12 \\ ∴ x & = & \pm \sqrt{y - 12} \end{array}$

So, $f^{- 1} (x) = - \sqrt{x - 12} or f^{- 1} (x) = \sqrt{x - 12}$

Therefore, the inverse of $f (x)$ is $f^{- 1} : [0, \infty) \to [12, \infty]$ defined by $f^{- 1} (x) = - \sqrt{x - 12} or f^{- 1} (x) = \sqrt{x - 12}$ .

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