Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #14 : Relations And Functions

Finding an inverse function.

Given


 

find

 

.

Possible Answers:

.

Correct answer:

.

Explanation:

Finding an inverse function is essentially finding the output in terms of the input.  To do this switch the independent and dependent variables.  First set

.

Then

Now, to switch the role of the variables solve for x in terms of y.

or

So, y is now the independent variable, and x is the dependent variable.  The final step is to write the inverse function in proper notation. 

Set  , and .

This gives

.

 

 

 

Example Question #15 : Relations And Functions

Which function below correctly translates  four units to the right and two units down?

Possible Answers:

Correct answer:

Explanation:

The translation  moves the graph of   units to the left if .  To see this, compare the graphs of  and .  A translation to the right must be in the form  if .

A vertical translation is up if .  We want to move the function down, so our translation subtracts two from .

Combining these, we get .

Example Question #51 : Pre Calculus

What is the domain of the following function?

Possible Answers:

-

Correct answer:

Explanation:

The denominator becomes zero whenever cos(x) becomes -1 and this happens when x is an odd multiple of .  To avoid division by zero, we exclude all these numbers. Therefore the domain is:

 

Example Question #2 : Angles

Solve for .

Question_2

(Figure not drawn to scale).

Possible Answers:

Correct answer:

Explanation:

The angles are supplementary, therefore, the sum of the angles must equal .

Example Question #7 : Angles

Are  and  supplementary angles?

Possible Answers:

No

Yes

Not enough information

Correct answer:

Yes

Explanation:

Since supplementary angles must add up to , the given angles are indeed supplementary.

Example Question #1 : Angles

Solve for and .

Question_3

(Figure not drawn to scale).

Possible Answers:

Correct answer:

Explanation:

The angles containing the variable  all reside along one line, therefore, their sum must be .

Because  and  are opposite angles, they must be equal.

Example Question #1 : Graphs Of Polynomial Functions

For what values of  will the given polynomial pass through the x-axis if plotted in Cartesian coordinates? 

Possible Answers:

None of the other answers

Correct answer:

Explanation:

One can remember that if the multiplicity of a zero is odd then it passes through the x-axis and if it's even then it 'bounces' off the x-axis. You can think about this analytically as well. What happens when we plug a number into our function just slightly above or below a zero with an even multiplicity? You find that the sign is always positive. Whereas a zero with an odd multiplicity will yield a positive on one side and a negative on the other. For zeros with odd multiplicity this alters the sign of our output and the function passes through the x-axis. Whereas the zero with even multiplicity will output a number with the same sign just above and below its zero, thus it 'bounces' off the x-axis.

Example Question #2 : Graphs Of Polynomial Functions

For this particular question we are restricting the domain of both  to nonnegative values, or the interval .

Let  and .  

For what values of  is ?

Possible Answers:

Correct answer:

Explanation:

The cubic function will increase more quickly than the quadratic, so the quadratic function must have a head start.  At , both functions evaluate to 8.  After than point, the cubic function will increase more quickly.

The domain was restricted to nonnegative values, so this interval is our only answer.

Example Question #51 : Pre Calculus

Triangle

What is the ?

Possible Answers:

 

Correct answer:

Explanation:

 

Example Question #2 : Graphs And Inverses Of Trigonometric Functions

Triangle

In the right triangle above, which of the following expressions gives the length of y?

Possible Answers:

Correct answer:

Explanation:

 is defined as the ratio of the adjacent side to the hypotenuse, or in this case . Solving for y gives the correct expression.

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