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College Algebra
Answer/Discussion to Practice Problems  
Tutorial 32B:
Inverse Functions



 

checkAnswer/Discussion to 1a
prob 1a1 and example 1a2


Let’s look at f(g(x)) first:


ad1a
*Insert the "value" of g inside the function of f
*Plug in the "value" of g wherever there is an x the function f

   

Since BOTH f(g(x)) AND g(f(x)) would have to equal x for them to be inverses of each other and f(g(x)) is not equal to x, then we can stop here and say without a doubt that they are NOT inverses of each other.




 

 

checkAnswer/Discussion to 1b
probl 1b1 and prob 1b2


Let’s look at f(g(x)) first:


ad1b1
*Insert the "value" of g inside the function of f

*Plug in the "value" of g wherever there is an x the function f

 

Next, let’s look at g(f(x)):


ad1b2
*Insert the "value" of f inside the function of g

*Plug in the "value" of f wherever there is an x the function g

 

Since f(g(x)) AND g(f(x)) BOTH came out to be x, this proves that the two functions are inverses of each other.


  


 

 

checkAnswer/Discussion to 2a
prob 2a



a) find the equation of f inverse


Step 1: Replace f(x) with y.

ad2a1
 


Step 2: Interchange x with y.

ada22
 


Step 3: Solve for y.

ada23
*Inverse of cube root is cubing both sides

*Inverse of -2 is + 2



Since this y is a function, it is the inverse of the original function.



Step 4: If f has an inverse function, replace y with f inverse.

ad2a4
 


b) graph f and f inverse

graph

See how the graphs of f and f inverse are mirrored images across the line y = x?  Note how the ordered pairs are interchanged … for example (2, 0) is a point on f where (0, 2) is a point on f inverse

If you need a review on graphing the cubic functions, feel free to go to College Algebra Tutorial 31: Graphs of Functions, Part I.



c) indicate the domain and range of f and f inverseusing interval notation


The domain of f is the same as the range of f inversewhich is ad2a6.

The range of f is the same as the domain of f inverse which is ad2a6 .

If you need a review on finding the domain and range of a function, feel free to go to College Algebra Tutorial 30: Introduction to Functions.

If you need a review on finding the domain and range of a graph of a function, feel free to go to College Algebra Tutorial 32: Graphs of Functions, Part II.


  


 

 

checkAnswer/Discussion to 2b
prob 2b


a) find the equation of f inverse


Step 1: Replace f(x) with y.

ad2b1
 


Step 2: Interchange x with y.

ad2b2
 


Step 3: Solve for y.

ad2b3
*Inverse of - 2 is + 2

*Inverse of squaring is to take the square root

Normally when we take the square root of both sides, there are two answers .. the + or – of the square root.  But, because it is only defined for greater than or equal to 0, then that means we can only use the positive square root.

If we had both + and – in front of the square root, then this would not be a function and hence there would not be an inverse. But, we do have a function here, so we can carry on ….

Since this y is a function, it is the inverse of the original function.



Step 4: If f has an inverse function, replace y with f inverse.

ad2b4
 


b) graph f and f inverse

graph

See how the graphs of f and f inverse are mirrored images across the line y = x?  Note how the ordered pairs are interchanged … for example (0, -2) is a point on f where (-2, 0) is a point on f inverse.

If you need a review on graphing the quadratic and square root functions, feel free to go to College Algebra Tutorial 31: Graphs of Functions, Part I.



c) indicate the domain and range of f and f inverseusing interval notation


The domain of f is the same as the range of f inverse which is ad2b6.

The range of f is the same as the domain of f inverse which is  ad2b7.

If you need a review on finding the domain and range of a function, feel free to go to College Algebra Tutorial 30: Introduction to Functions.

If you need a review on finding the domain and range of a graph of a function, feel free to go to College Algebra Tutorial 32: Graphs of Functions, Part II.


 

 

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Last revised on June 24, 2010 by Kim Seward.
All contents copyright (C) 2002 - 2010, WTAMU and Kim Seward. All rights reserved.