to view answer.
C, given the number of pencils you buy, n. So, in this case, you could write this as C(n) = some expression involving n. In other words, C is a function of n. This means that you “input” the number of pencils, n, and after performing some operations on it, you reach the specific “output”, which in this case is the cost, C. Sometimes the output of a function for an input is called the “value of the function” for that input.
Keep in mind that a defining aspect of a function is that for every input, there is exactly one output. This means that when you input a number into a function, you are guaranteed to get one (and only one) value as an output. This does not mean that every output is different. Two inputs can have the same output, but the important thing is that each input has only one output.
Let $$ A\left( t \right) = 6t - 8 $$ be a function.
(a) Which variable represents the input of the function? How do you know?
The input is represented by the letter t. We know that this is the input because it is the variable that we perform operations on to get an output value. It is the variable in parentheses, meaning that that the function is taken at that value, or, for that input.
(b) Which variable represents the output of the function? How do you know?
The output is represented by the letter A(t). A(t) is the value you get out when you input a value for t.
(c) What is the output when you input a value of 7 for t?
Here, we want the value of A when the input (t) is 7, and so, we want to find A(7). The function tells us what to do to our input value to get our output value. In this case, the function tells us to multiply by 6 and then subtract 8. We can write this:
$$
A\left( 7 \right) = \left( {6 \cdot 7} \right) - 8
You can see that wherever there was a t, we have placed the input value of t, which is 7. And, so we get:
$$
A\left( 7 \right) = 34
$$
Thus, an input value of 7 gives us an output value of 34.
(d) If the output value of the function is 52, what was the input?
In this case, we are given an output value, A(t) = 52, and asked to find the input value, t, which returns that output value. So, here, we simply substitute 52 in for A(t), and solve for t:
$$
52 = 6t - 8
$$
To solve for t, we want to find all the values of t that make the equation true. So, undo what has been done to t (get t by itself) to see which t values make the equation true:
Add 8 to both sides:
$$
60 = 6t
$$
And, divide by 6:
$$
10 = t
$$
Thus, the input value of 10 for t returns the output value of 52. We can write this:
$$
A\left( {10} \right) = 52
$$
NOTE: Although every input value of a function returns one and only one output value, it is not always true that every output value will return only one input value. More than one of the inputs may have the same output value. Thus, as you will see in Example 2, sometimes, when you solve for an input value given an output value, you get more than one answer.
Let $$
g\left( x \right) = 4x^2 - 6
$$.
(a) What is the output value when the input value, x, is $${\Large
{3 \over 2}}
$$
?
To solve this, substitute $$
{\Large{3 \over 2}}
$$in for x. Then, we have:
$$
g\left( {{\Large{3 \over 2}}} \right) = 4\left( {{\Large{3 \over 2}}} \right)^2 - 6
$$
$$
g\left( {{\Large{3 \over 2}}} \right) = 4\left( {{\Large{9 \over 4}}} \right) - 6
$$
$$
g\left( {{\Large{3 \over 2}}} \right) = 9 - 6
$$
So, finally:
$$
g\left( {{3 \over 2}} \right) = 3
$$
An input of $$ {\Large{3 \over 2}} returns an output of 3.
(b) If the output value of g is 42, what was the input?
Since the output is 42, replace the g(x) with 42 in the equation and solve for x:
$$
42 = 4x^2 - 6
$$
So, to solve for x, first add 6 to each side:
$$
48 = 4x^2
$$
Now, divide each side by 4:
$$
12 = x^2
$$
Now, for which value(s) of x does $$ x^2 = 12 $$ xcould also be the negative square root:
$$
x = \pm \sqrt {12}
$$
Which simplifies:
$$
x = \pm \sqrt {12} = \pm \sqrt {4 \cdot 3} = \pm \sqrt 4 \sqrt 3 = \pm 2\sqrt 3
$$
Thus, as you can see, x can have one of two different values. Both inputs, $$ x = 2\sqrt 3 $$ and $$ x = - 2\sqrt 3 give the same output of 42.
Let $$
f\left( y \right) = {\Large{{7y - 9} \over {14 + y}}}
$$
(a) What is the value of the function when y = $${\Large{1 \over 2}}
$$
? In other words, what is $$
f\left( {{\Large{1 \over 2}}} \right)
$$
?
To find the value of the function, input $$
{\Large{1 \over 2}}
$$ wherever y appears in the function:
$$
f\left( {\Large{1 \over 2}} \right) = {\Large{{7\left( {\LARGE{1 \over 2}} \right) - 9} \over {14 + \left( {\LARGE{1 \over 2}} \right)}}}
$$
Now we must simplify the numerator by finding a common denominator and then subtracting the fractions. (See adding fractions for extra help)
$$
f\left( {\Large{1 \over 2}} \right) = {\Large{{7\left( {\LARGE{1 \over 2}} \right) - 9} \over {14 + \left( {\LARGE{1 \over 2}} \right)}}} = {\Large{{\left( {\LARGE{7 \over 2}} \right) - \left( {\LARGE{18 \over 2}} \right)} \over {14 + \left( {{1 \over 2}} \right)}}} = {\Large{{ - \left( {\LARGE{11 \over 2}} \right)} \over {14 + \left( {\LARGE{1 \over 2}} \right)}}}
$$
Then, we add fractions in the denominator by, once again, finding a common denominator and then adding.:
$$
f\left( {\Large{1 \over 2}} \right) = {\Large{{ - \left( {{\LARGE{11 \over 2}}} \right)} \over {14 + \left( {\LARGE{1 \over 2}} \right)}}} = {\Large{{ - \left( {{\LARGE{11 \over 2}}} \right)} \over {\left( {{\LARGE{28 \over 2}}} \right) + \left( {\LARGE{1 \over 2}} \right)}}} = {\Large{{ - \left( {{\LARGE{11 \over 2}}} \right)} \over {\left( {{\LARGE{29 \over 2}}} \right)}}}
$$
Finally, we divide fractions, to get:
$$
f\left( {\Large{1 \over 2}} \right) = {\Large{{ - \left( {\LARGE{11 \over 2}} \right)} \over {\left( {\LARGE{29 \over 2}} \right)}}} = - \left( {\Large{11 \over 2}} \right)\left( {\Large{2 \over 29}} \right) = - {\Large{11 \over 29}}
$$
(b) What is the value of the function when the input is$$ y = 3a - 4 $$ ? In other words, what is $$ f\left( {3a - 4} \right) $$
This time, our input value is not just a number, but is the expression $$ \left( {3a - 4} \right) $$. However, even though it is more than just a number, it still works the same way. Input the expression $$ \left( {3a - 4} \right) $$ wherever y appears in the function:
$$
f\left( y \right) = {\Large{{7y - 9} \over {14 + y}}}
$$ and so: $$
f\left( {3a - 4} \right) = {\Large{{7\left( {3a - 4} \right) - 9} \over {14 + \left( {3a - 4} \right)}}}
$$
Now, use the distributive property and combine like terms to simplify:
$$
f\left( {3a - 4} \right) = {\Large{{7\left( {3a - 4} \right) - 9} \over {14 + \left( {3a - 4} \right)}}} = {\Large{{21a - 28 - 9} \over {14 + 3a - 4}}} = {\Large{{21a - 37} \over {3a + 10}}}
$$
NOTE: You cannot simplify this fraction any further. Do not cancel the 3 with the 21. Although they are both divisible by 3, there are constants (-37 and 10) added to terms 21a and 3a, respectively. Therefore, we cannot cancel them. We can only cancel numbers that are factored out of the whole expression.
Some practice problems to check your skills:
1. Let $$
f\left( x \right) = 8x - 7
$$
. What is the value of the function (the output) when we input 1.5 for x? In other words, what is $$
f\left( {1.5} \right)
$$
?
Think about it for a moment and then access this link to view answer.
2. Again, let$$ f\left( x \right) = 8x - 7 $$ . For what input do we get an output of 75?
Think about it for a moment and then access this link to view answer.
3. Let $$ A\left( t \right) = {\Large{{t^2 - 3t + 7} \over { - \left( {5t - 9} \right)}}} $$ . What is the value of the function when we input $$ - t $$ ? In other words, what is $$ A\left( { - t} \right) $$ ?
Think about it for a moment and then access this link to view answer.
4. Let $$ g\left( y \right) = {\Large{{6y - 8} \over {9 + y}}} $$ . What is $$ g\left( {2a - 3} \right) $$ ?
Think about it for a moment and then access this link to view answer.
5. Let $$ h\left( x \right) = 2x^2 - 8 $$ . What input value(s) give an output value of 24?
Think about it for a moment and then access this link to view answer.