Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
4. Arithmetic Series
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Exercise 60 Page 593

Consider two points on the given graph. Exchange the coordinates of those points. These two new points satisfy the equation of the inverse.

H

Practice makes perfect

When given a multiple choice exercise, it is a good idea to use the answer choices. In this case we can take points off the graph to help us. For simplicity, let's use the points (1,0) and (3,1).

Since these two points are on the graph of the given function, if we exchange their coordinates the resulting points must be on the graph of the inverse.

c|c Points on the graph & Points on the graph of the function & of the inverse [0.8em] ( 1, 0) and ( 3, 1) & ( 0, 1) and ( 1, 3) Now we can check which of the given equations is satisfied by the resulting points. Let's start with the point ( 0, 1).

Choice Equation Substitution Simplify
F y=3x 1 ? = 3 ( 0) 1≠ 0 *
G y=-3^(2x) 1 ? = -3^(2( 0)) 1≠ - 1 *
H y=3^x 1 ? = 3^0 1=1 ✓
I y=2^(3x) 1 ? = 2^(3( 0)) 1=1 ✓

The point ( 0, 1) does not satisfy the equations given in choices F and G. Therefore, none of these two functions is the inverse of the function whose graph is given. Let's finally use the point ( 1, 3) to determine which of the two remaining functions is the inverse.

Choice Equation Substitution Simplify
H y=3^x 3 ? = 3^1 3=3 ✓
I y=2^(3x) 3 ? = 2^(3( 1)) 3≠ 8 *

The only equation that is satisfied by the two points that lie on the graph of the inverse is the equation given in choice H. Therefore, the inverse of the function whose graph is given is y=3^x.