# Evaluating Functions

Q:  Find the following values of the function: f(x) = 3x + 1

a. f(0)       b. f(3)

c. f(-2)      d.f(-x)

e. -f(x)      f. f(x+2)

g. f(2x)     h. f(x+h)

A:  As I’ve mentioned before, a function is just a machine.  You put something in it (a variable, a number, etc…) and something else comes out (maybe a different number or a different variable).  So, let’s start:

(a) f(x) = 3x + 1

The red parts will always match.  So, f(7) = 3(7) + 1… Therefore, f(0) means to replace all of the x’s in the equation with 0’s:

f(0) = 3(0) + 1 = 0 + 1 = 1

(b)

f(3) = 3(3) + 1 = 9 + 1 = 10

Again, see how the “3” just replaced the x?

(c)

f(-2) = 3(-2) + 1 = -6 + 1 = -5

(d) Find f(-x).  This is a little more confusing to look at, but no different than before.  The “-x” just replaces all of the x’s:

f(-x) = 3(-x) + 1 = -3x + 1

f(x) = 3x + 1

Now, add a negative in front of both sides:

-f(x) = -(3x + 1)

-f(x) = -3x – 1

(f) Back to just plugging in to the function:

f(x + 2) = 3(x + 2) + 3 = 3x + 6 + 3 = 3x + 9

(g) Plug in again:

f(2x) = 3(2x) + 1 = 6x + 1

(h) More with the plugging in:

f(x + h) = 3(x + h) + 1

f(x + h) = 3x + 3h + 1

Wow.  That was all of them.

Remember…. A function is a set of rules… f(x) is read as “f of x”: meaning that the function f depends on the input x.

f(3) is read as “f of 3” and it means to plug “3” into the function.

Please notice the difference between the two:

f(x + h) and f(x) + h

Notice that the problem in read has (x + h) plugged into the function…. Therefore, using our function f(x) = 3x + 1, that would make:

f(x + h) = 3(x + h) + 1

Notice that the (x + h) went into the function replacing the original x.

Now, consider the problem in blue : f(x) + h.  This is the function f(x) with an h added to the end. So, if f(x) = 3x + 1, then:

f(x) + h = (3x + 1) + h

See the difference?  This is a very important concept to understand about functions.