Example 1 Determine which of the
following equations are functions and which are not functions.
(a) y = 5x +1
(b) y = x2 +1
(c) y2 = x +1
(d) x2 + y2 = 4
Solution
The definition of function is saying is that if we take all
possible values of x and plug them into the equation and solve for y we
will get exactly one value for each value of x. At this stage of the game
it can be pretty difficult to actually show that an equation is a function so
we’ll mostly talk our way through it. On the other hand it’s often quite easy
to show that an equation isn’t a function.
(a) y = 5x +1
So,
we need to show that no matter what x we plug into the equation and
solve for y we will only get a single value of y. Note as well
that the value of y will probably be different for each value of x, although it doesn’t have to be.
Let’s
start this off by plugging in some values of x and see what happens.
x = – 4
: y
= 5( – 4) + 1 = – 20 + 1 = – 19
x = 0 y = 5(0) + 1 = 0 + 1 = 1
x = 10 y = 5(10) + 1 = 50 + 1 = 51
So,
for each of these value of x we got a single value of y out of
the equation. Now, this isn’t sufficient to claim that this is a function. In
order to officially prove that this is a function we need to show that this
will work no matter which value of x we plug into the equation.
Of
course we can’t plug all possible value of x into the equation. That
just isn’t physically possible. However, let’s go back and look at the ones
that we did plug in. For each x, upon plugging in, we first multiplied
the x by 5 and then added 1 onto it. Now, if we multiply a number by 5
we will get a single value from the multiplication. Likewise, we will only get
a single value if we add 1 onto a number. Therefore, it seems plausible that
based on the operations involved with plugging x into the equation that
we will only get a single value of y out of the equation.
So, this equation is a function.
(b) y = x2 +1
Again, let’s plug in a couple of values of x and solve for y
to see what happens.
x
= – 1 :
y = (– 1)2 + 1 = 1 + 1 = 2
x
= 3 : y =
(3)2 + 1 = 9 + 1 = 10
Now, let’s think a little bit about what we were doing with the
evaluations. First we squared the value of x that we plugged in. When we
square a number there will only be one possible value.We then add 1 onto this,
but again, this will yield a single value.
So, it seems like this equation is also a function.
Note that it is okay to get the same y value for different x’s.
For example,
x = – 3 : y = (– 3)2 + 1 = 9 +1 =10
We just can’t get more than one y out of the
equation after we plug in the x.
(c) y2 = x +1
As
we’ve done with the previous two equations let’s plug in a couple of value of x,
solve for yand see what we get.
x = 3 : y2
= 3 + 1 = 4 ⇒ y
= ±2
x = − 1 : y2 =
− 1 + 1 = 0 ⇒ y
= 0
x = 10 : y2
= 10 + 1 = 11 ⇒ y
= ±√11
Now, remember that we’re solving for y and so that means
that in the first and last case above we will actually get two different y values
out of the x and so this equation is NOT a function. Note that we can have values of x that will yield a single y
as we’ve seen above, but that doesn’t matter. If even one value of x yields more than one value
of y upon solving the equation will not be a function.
What this really means is that we didn’t need to go any farther
than the first evaluation, since that gave multiple values of y.
(d) x2 + y2 = 4
With this case we’ll use the lesson learned in the previous part
and see if we can find a value of x that will give more than one value
of y upon solving. Because we’ve got a y2 in the problem this shouldn’t
be too hard to do since solving will eventually mean using the square root property which will give more than one value of y.
x = 0 : 02
+ y2 = 4 ⇒ y2 = 4 ⇒ y = ± 2
So, this equation is not a function. Recall, that from the
previous section this is the equation of a circle. Circles are never functions.
Example 2 Given f (x) = x2 − 2x
+8 and g (x) = x + 6 evaluate each of the following.
(a) f (3) and g (3)
(b) f (−10) and g (−10)
(c) f (0)
(d) f (t )
(e) f (t +1) and f (x +1)
(f) f (x3 )
(g) g (x2 −5)
Solution
(a) f (3) and g (3)
Okay
we’ve got two function evaluations to do here and we’ve also got two functions
so we’re going to need to decide which function to use for the evaluations. The
key here is to notice the letter that is in front of the parenthesis. For f (3) we will use the
function f (x) and for g (3) we will use g
(x). In other
words, we just need to make sure that the variables match up.
Here
are the evaluations for this part.
f (3) = (3)2 – 2(3) + 8 = 9 – 6 + 8 =
11
(b) f (−10) and g (−10)
This one is pretty much the same as the previous part with one
exception that we’ll touch on when we reach that point. Here are the
evaluations.
f (−10)
= (−10)2 − 2 (−10) + 8 = 100 + 20 + 8 = 128
Make sure that you deal with the negative signs properly here. Now
the second one.
We’ve now reached the difference. Recall that when we first
started talking about the definition of functions we stated that we were only going to deal with real
numbers. In other words, we only plug in real numbers and we only want real numbers back out
as answers. So, since we would get a complex number out of this we can’t plug -10 into this
function.
(c) f (0)
Not much to this one.
f (0) = (0)2
− 2 (0) + 8 = 8
Again, don’t forget that this isn’t multiplication! For some
reason students like to think of this one as multiplication and get an answer of zero. Be careful.
(d) f (t )
The
rest of these evaluations are now going to be a little different. As this one
shows we don’t need to just have numbers in the parenthesis. However,
evaluation works in exactly the same way.
We plug into the x’s on the right side of the equal sign whatever is in
the parenthesis. In this case that means that we plug in t for all the x’s.
Here
is this evaluation.
f (t ) = t2 − 2t +8
Note
that in this case this is pretty much the same thing as our original function,
except this time we’re using t as a variable.
(e) f (t +1) and f ( x +1)
Now,
let’s get a little more complicated, or at least they appear to be more
complicated. Things aren’t as bad as they may appear however. We’ll evaluate f
(t + 1) first. This one
works exactly the same as the previous part did. All the x’s on the left
will get replaced with t + 1. We will have some simplification to do as well after the
substitution.
f (t
+ 1) = (t + 1)2 –
2(t + 1) + 8
= t2 + 2t + 1 – 2t
– 2 + 8
= t2 + 7
Be
careful with parenthesis in these kinds of evaluations. It is easy to mess up
with them.
Now,
let’s take a look at f (x +1). With the
exception of the x this is identical to f (t + 1) and so
it works exactly the same way.
f
(x + 1) = (x + 1)2 – 2(x + 1) + 8
= x2 + 2x + 1 – 2x
– 2 + 8
= x2 + 7
Do not get excited about the fact that we reused x’s in the
evaluation here. In many places where we will be doing this in later sections
there will be x’s here and so you will need to get used to seeing that.
(f) f (x3)
Again, don’t get excited about the x’s in the parenthesis
here. Just evaluate it as if it were a number.
f
(x3) = (x3)2
– 2(x3) + 8
= x6 – 2x3
+ 8
(g) g (x2 −5)
One more evaluation and this time we’ll use the other function.
Example 3 Given,
evaluate each of the following.
(a) g (−6)
(b) g (−4)
(c) g (1)
(d) g (15)
(e) g (21)
Solution
Before starting the evaluations here let’s notice that we’re using
different letters for the function and variable than the ones that we’ve used
to this point. That won’t change how the evaluation works. Do not get so locked
into seeing f for the function and x for the variable that you
can’t do any problem that doesn’t have those letters.
Now,
to do each of these evaluations the first thing that we need to do is determine
which inequality the number satisfies, and it will only satisfy a single
inequality. When we determine which inequality the number satisfies we use the
equation associated with that inequality.
So, let’s do some evaluations.
(a) g (−6)
In this case -6 satisfies the top inequality and so we’ll use the
top equation for this evaluation.
g (−6) = 3 (−6) + 4 = 112
(b) g (−4)
Now we’ll need to be a little careful with this one since -4 shows
up in two of the inequalities. However, it only satisfies the top inequality
and so we will once again use the top function for the evaluation.
g (−4) = 3 (−4) + 4 = 52
(c) g (1)
In this case the number, 1, satisfies the middle inequality and so
we’ll use the middle equation for the evaluation. This evaluation often causes
problems for students despite the fact that it’s actually one of the easiest
evaluations we’ll ever do. We know that we evaluate functions/equations by
plugging in the number for the variable. In this case there are no variables.
That isn’t a problem. Since there aren’t any variables it just means that we
don’t actually plug in anything and we get the following,
g (1) =10
(d) g (15)
Again, like with the second part we need to be a little careful
with this one. In this case the number satisfies the middle inequality since
that is the one with the equal sign in it. Then like the previous part we just
get,
g (15) =10
Don’t get excited about the fact that the previous two evaluations
were the same value. This will happen on occasion.
(e) g (21)
For the final evaluation in this example the number satisfies the
bottom inequality and so we’ll use the bottom equation for the evaluation.
g (21) =1− 6(21) = −125
Example 4 Determine the domain of
each of the following functions.
Solution
The domains for these functions are all the values of x for
which we don’t have division by zero or the square root of a negative number.
If we remember these two ideas finding the domains will be pretty easy.
So, in this case there are no square roots so we don’t need to
worry about the square root of a negative number. There is however a
possibility that we’ll have a division by zero error. To determine if we will
we’ll need to set the denominator equal to zero and solve.
x2 + 3x − 10 = (x + 5)(x − 2) = 0 x = −5, x
= 2
So, we will get division by zero if we plug in x = −5
or x = 2 . That means that we’ll need to avoid those two numbers.
However, all the other values of x will work since they don’t give division
by zero. The domain is then,
Domain : All real numbers except x = −5
and x = 2
In this case we won’t have division by zero problems since we
don’t have any fractions. We do have a square root in the problem and so we’ll
need to worry about taking the square root of a negative numbers.
This one is going to work a little differently from the previous
part. In that part we determined the value(s) of x to avoid. In this case it will be just as
easy to directly get the domain. To avoid square roots of negative numbers all that we need to do is require
that.
5 − 3x ≥ 0
This is a fairly simple linear inequality that we should be able
to solve at this point.
5 ≥ 3x ⇒ 
The domain of this function is then,
Domain :
In this case we’ve got a fraction, but notice that the denominator will never be zero for any real number since x2 is guaranteed to be positive or zero and adding 4 onto this will mean that the denominator is always at least 4. In other words, the denominator won’t ever be zero. So, all we need to do then is worry about the square root in the numerator.
To
do this we’ll require,
7x
+ 8 ≥ 0
7x
≥ − 8
Now,
we can actually plug in any value of x into the denominator, however,
since we’ve got the square root in the numerator we’ll have to make sure that
all x’s satisfy the inequality above to avoid problems. Therefore, the
domain of this function is
Domain :
In this final part we’ve got both a square root and division by zero to worry about. Let’s take care of the square root first since this will probably put the largest restriction on the values of x. So, to keep the square root happy (i.e. no square root of negative numbers) we’ll need to require that,
10x – 5 ≥ 0
So,
at the least we’ll need to require thatin
order to avoid problems with the square root.
in
order to avoid problems with the square root.
Now,
let’s see if we have any division by zero problems. Again, to do this simply
set the denominator
equal to zero and solve.
x2 −16 = (x − 4)( x + 4) = 0 ⇒ x = −4, x = 4
Now,
notice that x = −4 doesn’t
satisfy the inequality we need for the square root and so that value of x has
already been excluded by the square root. On the other hand x = 4 does satisfy the
inequality. This means that it is okay to plug x = 4 into the
square root, however, since it would give division by zero we will need to
avoid it.
The
domain for this function is then,
Domain :except x = 4.
except x = 4.Sumber
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