College Algebra Tutorial 30

College Algebra
Answer/Discussion to Practice Problems
Tutorial 30: Introduction to Functions

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Answer/Discussion to 1a

{(1, 5), (2, 6), (3, 7), (4, 8)}

Is this a function or not?
We need to ask ourselves, does every first element (or input) correspond with EXACTLY ONE second element (or output)? In this case, the answer is yes. 1 only goes with 5, 2 only goes with 6, 3 only goes with 7, and 4 only goes with 8.

So, this relation would be an example of a function.

Domain
We need to find the set of all input values. In terms of ordered pairs, that correlates with the first component of each one. So, what do you get for the domain?

If you got {1, 2, 3, 4}, you are correct!

Range
We need to find the set of all output values. In terms of ordered pairs, that correlates with the second component of each one. SO, what do you get for the range?

If you got {5, 6, 7, 8}, you are absolutely right!

(return to problem 1a)

Answer/Discussion to 1b

{(-1, 2), (-1, 3), (2, 4)}

Is this a function or not?
We need to ask ourselves, does every first element (or input) correspond with EXACTLY ONE second element (or output)? In this case, the answer is no. The input value of -1 goes with two output values, 2 and 3. It only takes one input value to associate with more than one output value to be invalid as a function.

So, this relation would not be an example of a function.

Domain
We need to find the set of all input values. In terms of ordered pairs, that correlates with the first component of each one. So, what do you get for the domain?

If you got {-1, 2}, you are correct!

Note that if any value repeats, we only need to list it one time.

Range
We need to find the set of all output values. In terms of ordered pairs, that correlates with the second component of each one. So, what do you get for the range?

If you got {2, 3, 4}, you are absolutely right!

(return to problem 1b)

Answer/Discussion to 2a

To check if y is a function of x, we need to solve for y and then check to see if there is only one output for every input.

*Inverse of add. x squared is sub. x squared

*Inverse of squaring is taking the sq. root
*Solved for y

At this point we ask ourselves, would we get one value for y if you plug in any value for x?

If you answered no, you are correct.

For example, if our input value x is 0, then our output value y could either be 1 or -1. Note that I could have picked an infinite number of examples like this one. You only need to show one example where the input value is associated with more than one output value to disqualify it from being a function.

This means that at least one input value is associated with more than one output value, so by definition, y is not a function of x.

(return to problem 2a)

Answer/Discussion to 2b

To check if y is a function of x, we need to solve for y first and then check to see if there is only one output for every input.

*Inverse of add. 2x is sub. 2x
*Solved for y

At this point we ask ourselves, would we get one value for y if you plug in any value for x?

If you answered yes, you are right on. For example, if we plugged in a 1 for x, then y would only equal one value, 5. Note that ANY value you would plug in for x would produce only one value for y.

Note that since it is solved for y, y is our output value and x is our input value.

Since our answer to that question is yes, that means by definition, y is a function of x.

(return to problem 2b)

Answer/Discussion to 3a

Again, think of functional notation as a fancy assignment statement. For example, when we are looking for f(0), we are going to plug in 0 for the value of x in our function f and so forth.

*Plug in -5 for x and evaluate

*Plug in 2 for x and evaluate

So our answers are f(-5) = 6 and f(2) = 1.

(return to problem 3a)

Answer/Discussion to 4a

f(a):

Don't let the fact that we need to plug in an a throw you. You plug it into the function just like you do a number. Everywhere you have an x in your function, replace it with an a.

*Plug in a for x

f(a + h)

Don't let the fact that we need to plug in the expression a + h throw you. You plug it into the function just like you do a number. Everywhere you have an x in your function, replace it with an a + h.

*Plug in a + h for x and evaluate

Putting it all together we get:

*Plug in f( a + h) and f(a) found above

*Combine like terms

*Divide out h from every term

(return to problem 4a)

Answer/Discussion to 5a

problem 5a

To find g(-1), we need to go to the piece of the function that x = -1 would be under, which would be the first one where x< 0.

*This piece of the function is used when x < 0
*This is a constant function, it is always equal to 7

To find g(0), we need to go to the piece of the function that x = 0 would be under, which would be the first one where x < 0.

*This piece of the function is used when x < 0
*This is a constant function, it is always equal to 7

To find g(4), we need to go to the piece of the function that x = 4 would be under, which would be the last one where x > 0.

*This piece of the function is used when x > 0

*Plug in 4 for x

(return to problem 5a)

Answer/Discussion to 6a

Since we can plug in any value for x (input variable) and get a real number answer for our function (output variable), then there are no restrictions. Therefore the domain is all real numbers.

(return to problem 6a)

Answer/Discussion to 6b

Our restriction here is that the denominator of a fraction can never be equal to 0. So to find our domain, we want to set the denominator equal to 0 and restrict those values.

*The den. cannot equal to 0
*Factoring to help "solve"
*3 and -1 are restricted values

Our domain would be all real numbers except 3 and -1, because 3 and -1 both make the denominator equal to 0, which would not give us a real number answer for our function.

(return to problem 6b)

Answer/Discussion to 6c

This time we do have a restriction on our domain. We can only take the square root of a number greater than or equal to 0. So we must set the radicand (the inside of the radical) greater than or equal to 0.

*Inv. of add. 1 is sub. 1

*Inv. of mult. 2 is div. 2

*This is our domain

Only values of x that are greater than or equal to -1/2 make the original radicand greater than or equal to 0, so x > -1/2 is our domain for this function.

(return to problem 6c)

WTAMU > Virtual Math Lab > College Algebra > Tutorial 30: Introduction to Functions