Exercise 127 Page 250

To graph the function, we will start by making a table of values. Since we cannot take the square root of a negative number, x must be greater than or equal to 1. Any value less than that, would produce a negative argument under the square root.

|c|c|c| [-1em] x & 2+sqrt(x-1) & f(x) [0.2em] [-1em] 1 & 2+sqrt(1-1) & 2 [0.2em] [-1em] 2 & 2+sqrt(2-1) & 3 [0.2em] [-1em] 5 & 2+sqrt(5-1) & 4 [0.2em] [-1em] 10 & 2+sqrt(10-1) & 5 [0.2em]Let's mark these points in a coordinate plane and draw our graph.

Examining the graph, we notice that the domain is all values greater than or equal to 1. Additionally, the range is all values greater than or equal to 2. Domain f(x): & x ≥ 1 Range f(x): & y ≥ 2

To find the inverse function we must first switch x and f(x) in the equation.

Function:& f(x)=1+sqrt(x+5) Switch:& x=1+sqrt(f(x)+5) Let's now solve this equation for f(x).

x=2+sqrt(f(x)-1)

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Solve for f(x)

RearrangeEqn

Rearrange equation

2+sqrt(f(x)-1)=x

SubEqn

LHS-2=RHS-2

sqrt(f(x)-1)=x-2

RaiseEqn

LHS^2=RHS^2

f(x)-1=(x-2)^2

AddEqn

LHS+1=RHS+1

f(x)=(x-2)^2+1

Every point that falls on f(x) has a corresponding point on its inverse where the x- and y-coordinates of the points on f(x) have switched position. Using the table of values from Part A, we can identify four points on the inverse.

|c|c| [-1em] f(x) & f^(-1)(x) [0.2em] [-1em] (1,2) & (2,1) [0.2em] [-1em] (2,3) & (3,2) [0.2em] [-1em] (5,4) & (4,5) [0.2em] [-1em] (10,5) & (5,10) [0.2em]Just like we must switch position of the x- and y-coordinate of the points on f(x) to get some points on f^(-1)(x), we must also switch the domain and range. In other words, the domain for the function becomes the range of the inverse and vice verse. Domain f^(-1)(x): & x ≥ 2 Range f^(-1)(x): & y ≥ 1 Now we can graph the inverse.

Here, we have two different composite functions. In the first one, the function has been substituted into the inverse. In the second, the inverse has been substituted into the function.

f^(-1)( f(x))&=( (2+sqrt(x-1))-2)^2+1 [0.5em] f(f^(-1)(x))&= 2+sqrt(((x-2)^2+1)-1) Before we calculate the composite functions for x=5 we will simplify their right-hand sides.

f^(-1)(f(x))=((2+sqrt(x-1))-2)^2+1

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Simplify right-hand side

RemovePar

Remove parentheses

f^(-1)(f(x))=(2+sqrt(x-1)-2)^2+1

SubTerm

Subtract term

f^(-1)(f(x))=(sqrt(x-1))^2+1

PowSqrt

( sqrt(a) )^2 = a

f^(-1)(f(x))=x-1+1