Core Connections Integrated III, 2015
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Core Connections Integrated III, 2015 View details
1. Section 6.1
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Exercise 22 Page 275

The graphs of a function and its inverse are symmetric to each other with respect to the line y=x.

Equation: f^(-1)(x) = sqrt(2x)-1
Graph:

Practice makes perfect

To graph the given function, we can make a table of values to find the points on the graph.

x 1/2(x+1)^3 f(x)=1/2(x+1)^3
-3 1/2( -3+1)^3 -4
-2 1/2( -2+1)^3 -1/2
-1 1/2( -1+1)^3 0
0 1/2( 0+1)^3 1/2
1 1/2( 1+1)^3 4

Let's plot and connect the obtained points.

Now, we can graph the inverse of the function by reflecting the graph across y=x. This means that we should exchange the x and y coordinates of the points that we found.

Points Reflection across y=x
( -3, -4) ( -4, -3)
( -2, -1/2) ( -1/2, -2)
( -1, 0) ( 0, -1)
( 0, 1/2) ( 1/2, 0)
( 1, 4) ( 4, 1)

Once again, let's add the points to the graph and connect them.

Now we should move on to finding the equation of the inverse. Before we do it we will rewrite the given function as an equation relating x and y. f(x)=1/2(x+1)^3 ⇒ y=1/2(x+1)^3 To algebraically determine the inverse of the given relation, we exchange x and y and solve for y. c|c Given Equation & Inverse Equation [0.8em] y=1/2( x+1)^3 & x=1/2( y+1)^3 The result of isolating y in the new equation will be the inverse of the given function.
x=1/2(y+1)^3
Solve for y
2x=(y+1)^3
sqrt(2x) = sqrt((y+1)^3)

( sqrt(a) )^n = a

sqrt(2x)=y+1
sqrt(2x)-1=y
y=sqrt(2x)-1
Now that we have isolated y, we have found the equation of the inverse of the given function. y=sqrt(2x)-1 ⇒ f^(-1)(x) = sqrt(2x)-1