Chapter 1, Problem 98CR

(a)

To determine

To prove: f and g are inverse function algebraically.

Explanation of Solution

Given data:

  $\begin{array}{l}f(x)=\sqrt{x+1}\\ g(x)=x^2+1,x\ge 0\end{array}$

Calculation:

Algebraically

Let f(x) be y

So,

  $\begin{array}{l}f(x)=\sqrt{x+1}\\ y=\sqrt{x+1}\\ \text{for inverse function,}\\ x+1=y^2\\ x=y^2-1\\ f^{-1}(x)=x^2-1\\ f^{-1}(x)=g(x)\end{array}$

g(x) is inverse function of f(x).

(b)

To determine

To prove: f and g are inverse function graphically.

Explanation of Solution

Given data:

  $\begin{array}{l}f(x)=\sqrt{x+1}\\ g(x)=x^2+1,x\ge 0\end{array}$

Graph:

Graphically:

  

As the function g(x) is the reflection of the function f(x). both functions are inverse s of each other.

So g(x) is inverse function of f(x).

(c)

To determine

To prove: f and g are inverse function numerically.

Explanation of Solution

Given data:

  $\begin{array}{l}f(x)=\sqrt{x+1}\\ g(x)=x^2+1,x\ge 0\end{array}$

Calculation:

Numerically:

  $f(x)=\sqrt{x+1}\text{ ; }g(x)=x^2-1,x\ge 0$

Domain of f(x): $x>-1$ Domain of g(x): $x>0$

Range of f(x): $x>0$ Range of g(x): $x>-1$

So, g(x) is an inverse of f(x) as domain of f(x) is equal to range of g(x).