## Tuesday, November 27, 2007

### Inverse Trigonometric Functions. Inverse

The three trigonometric functions studied in this tutorial are: arcsin(x), arccos(x) and arctan(x).

The exploration is carried out by analyzing the graph of the function and the graph of its inverse. The domain and range of each of the above functions are also explored. Follow the steps in the tutorial below.

Several questions with detailed solutions as well as exercises with answers on one to one functions are presented.

From the definition of one-to-one functions we can write that a given function f(x) is one-to-one

if A is not equal to B then f(A) is not equal f(B)

where A and B are any values of the variable x in the domain of function f.

The contrapositive of the above definition is as follows:

if f(A) = f(B) then A = B

where A and B are any values of x included in the domain of f. We will use this contrapositive of the definition of one to one functions to find out whether a given function is a one to one.

Question 1: Is function f defined by

f = {(1 , 2),(3 , 4),(5 , 6),(8 , 6),(10 , -1)}

,

a one to one function?

Solution to Question 1:

Two different values in the domain, namely 5 and 6, have the same output, hence function f is not a one to one function.

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Question 2: Is function g defined by

g = {(-1 , 2),(0 , 4),(2 , -4),(5 , 6),(10 , 0)}

,

a one to one function?

Solution to Question 2:

Consider any two different values in the domain of function g and check that their corresponding output are different. Hence function g is a one to one function

Question 3: Is function f given by

f(x) = -x 3 + 3 x 2 - 2

,

a one to one function?

Solution to Question 3:

A graph and the horizontal line test can help to answer the above question.

Question 4: Show that all linear functions of the form

f(x) = a x + b

,

where a and b are real numbers such that a not equal to zero, are one to one functions.

Solution to Question 4:

We start with f(A) = f(B) and show that this leads to a = b

a(A) + b = a(B) + b

Add -b to both sides of the equation to obtain

a(A) = a(B)

Divide both sides by a since it is not equal to zero to obtain

A = B

Since we have proved that f(A) = f(B) leads to A = B then all linear functions of the form f(x) = a x + b are one-to-one functions.

A = B

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Question 5: Show that all functions of the form

f(x) = a (x - h) 2 + k , for x >= h

,

where a, h and k are real numbers such that a not equal to zero, are one to one functions.

Solution to Question 5:

We start with f(A) = f(B)

a (A - h) 2 + k = a (B - h) 2 + k

Add -k to both sides of the equation to obtain

a (A - h) 2 = a (B - h) 2

Divide both sides by a since it not equal to 0

(A - h) 2 = (B - h) 2

The above equation leads to two other equations

(A - h) = (B - h) or (A - h) = - (B - h)

The first equation leads

A = B

Let us examine the second equation. The domain of f is all values of x such that x >= h. This leads to x - h >= 0 which in turn leads to A - h >= 0 and B - h >= 0 which means the second equation (A - h) = - (B - h) does not have a solution.

--------------------------------------------------------------------------------

Question 6: Is function f given by

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

,

a one to one function?

Solution to Question 6:

It is easy to find two values of x that correspond to the same value of the function. f(0) = 1 / 4 and f(5) = 1 / 4. For two different values in the domain of f correspond one same value of the range and therefore function f is not a one to one.

--------------------------------------------------------------------------------

Question 7: Show that all the rational functions of the form

f(x) = 1 / (a x + b)

where a, and b are real numbers such that a not equal to zero, are one to one functions.

Solution to Question 7:

Let us write an equation starting with f(A) = f(B)

1 / (a A + b) = 1 / (a B + b)

Multiply both sides of the equation by (a A + b)(a B + b) and simplify

a B + b = a A + b

Add -b to both sides

a B = a A

Divide both sides by a to obtain

B = A

The given functions are one to one functions

The exploration is carried out by analyzing the graph of the function and the graph of its inverse. The domain and range of each of the above functions are also explored. Follow the steps in the tutorial below.

Several questions with detailed solutions as well as exercises with answers on one to one functions are presented.

From the definition of one-to-one functions we can write that a given function f(x) is one-to-one

if A is not equal to B then f(A) is not equal f(B)

where A and B are any values of the variable x in the domain of function f.

The contrapositive of the above definition is as follows:

if f(A) = f(B) then A = B

where A and B are any values of x included in the domain of f. We will use this contrapositive of the definition of one to one functions to find out whether a given function is a one to one.

Question 1: Is function f defined by

f = {(1 , 2),(3 , 4),(5 , 6),(8 , 6),(10 , -1)}

,

a one to one function?

Solution to Question 1:

Two different values in the domain, namely 5 and 6, have the same output, hence function f is not a one to one function.

--------------------------------------------------------------------------------

Question 2: Is function g defined by

g = {(-1 , 2),(0 , 4),(2 , -4),(5 , 6),(10 , 0)}

,

a one to one function?

Solution to Question 2:

Consider any two different values in the domain of function g and check that their corresponding output are different. Hence function g is a one to one function

Question 3: Is function f given by

f(x) = -x 3 + 3 x 2 - 2

,

a one to one function?

Solution to Question 3:

A graph and the horizontal line test can help to answer the above question.

Question 4: Show that all linear functions of the form

f(x) = a x + b

,

where a and b are real numbers such that a not equal to zero, are one to one functions.

Solution to Question 4:

We start with f(A) = f(B) and show that this leads to a = b

a(A) + b = a(B) + b

Add -b to both sides of the equation to obtain

a(A) = a(B)

Divide both sides by a since it is not equal to zero to obtain

A = B

Since we have proved that f(A) = f(B) leads to A = B then all linear functions of the form f(x) = a x + b are one-to-one functions.

A = B

--------------------------------------------------------------------------------

Question 5: Show that all functions of the form

f(x) = a (x - h) 2 + k , for x >= h

,

where a, h and k are real numbers such that a not equal to zero, are one to one functions.

Solution to Question 5:

We start with f(A) = f(B)

a (A - h) 2 + k = a (B - h) 2 + k

Add -k to both sides of the equation to obtain

a (A - h) 2 = a (B - h) 2

Divide both sides by a since it not equal to 0

(A - h) 2 = (B - h) 2

The above equation leads to two other equations

(A - h) = (B - h) or (A - h) = - (B - h)

The first equation leads

A = B

Let us examine the second equation. The domain of f is all values of x such that x >= h. This leads to x - h >= 0 which in turn leads to A - h >= 0 and B - h >= 0 which means the second equation (A - h) = - (B - h) does not have a solution.

--------------------------------------------------------------------------------

Question 6: Is function f given by

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

,

a one to one function?

Solution to Question 6:

It is easy to find two values of x that correspond to the same value of the function. f(0) = 1 / 4 and f(5) = 1 / 4. For two different values in the domain of f correspond one same value of the range and therefore function f is not a one to one.

--------------------------------------------------------------------------------

Question 7: Show that all the rational functions of the form

f(x) = 1 / (a x + b)

where a, and b are real numbers such that a not equal to zero, are one to one functions.

Solution to Question 7:

Let us write an equation starting with f(A) = f(B)

1 / (a A + b) = 1 / (a B + b)

Multiply both sides of the equation by (a A + b)(a B + b) and simplify

a B + b = a A + b

Add -b to both sides

a B = a A

Divide both sides by a to obtain

B = A

The given functions are one to one functions