1 Inverse Functions
We know what it means for a relation to be a function.
Every input maps to only one output, it passes the
vertical line test. But not every function has an
inverse. A function has no inverse if there are outputs
that are mapped to by different inputs such as in the
case of y = x2.
If a function has an inverse then we can determine the input if we know the output. For example if
the function
f (x) = x3 − 7
gave an output of 20 what was the input? What about
if the output was 5.167? How are you solving this?
The function you are using is the inverse, f −1(x).
Find it algebraically.
Graphically, a function has an inverse if it passes
the horizontal line test. A function that has an
inverse is said to be a one − to − one function. That
means every one input goes to only one output. In
math we can write
x1 6= x2 ⇒ f (x1) 6= f (x2)
Prove the example above is a one − to − one
function.
Begin by assuming that x1 6= x2 but f (x1) = f (x2)
and prove by contradiction.
Of course that makes good sense because we
already found the inverse and we can draw the graph
and show it passes the horizontal line test.
So let’s say a function, y = f (x) is one-to-one
with domain A and range B. The inverse function,
f −1(y) = x, has domain B and range A.
Notice how the range and domain switch places.
So what’s
f −1(f (a)) =
f (f −1(b)) =
and where does the a and b come from?
So we know how to algebraically get f −1(x) from
f (x) but how do we get the graph of the inverse from
the original graph?
As an example of the technique, sketch the graph
of y = ex and then find its inverse.
2 Logarithms
The logarithm function, y = logb(x) is defined for all
b > 0, b 6= 1, where b is the base. The graph looks like
1
0.5
1
x
1.5
2
2.5
3
0
-1
-2
-3
for all b’s. We can see that the range is all the
reals and the domain is (0, ∞). This introduces a new
complication into our “find the domain of the function”
questions. Before this we worried about not dividing
by zero and not taking the square root of a negative
number. We also have to make sure not to take the
log of zero or a negative number.
So what’s logarithm? Well, when I write logb(x) I
am saying “What do I have to raise b to, to get x?”. In
other words
y = logb(x)
is equivalent to
by = x
If there is no base written, it is assumed to be
base 10, like on your calculator. That is, log(100) = 2.
Example
Q?
Evaluate log2(16).
A.
What number do I have to raise 2 to to get
16?, Well 4, since 24 = 16
Example
1
Q?
Evaluate log3( 27
).
A.
What number do I have to raise 3 to to get
1
27 ?, Well it must be negative to make the fraction “1
1
over” and note that 3 3 = 27, therefore log3( 27
) = −3.
That is 3−3 = 27.
Example
Q?
Evaluate log29(29).
A.
What number do I have to raise 29 to to
get 29?, Well 1, since 291 = 29. This would have been
the answer for any base, not just 29.
Example
Q?
Evaluate log17(1).
A.
What number do I have to raise 17 to to
get 1?, Well 0, since 170 = 1. This would have been
the answer for any base, not just 17.
2.1 The Natural Logarithm, ln(x)
The logarithmic function y = loge(x) occurs so often
that it was given its own symbol, ln(x). That is
ln(x) = loge(x). The rules for regular log’s apply for
ln(x) also. The function y = ln(x) is the inverse
function to y = ex, that is, they undo each other.
ln(ex) = x
eln(x) = x
We will use this fact to solve for x. Try taking the
ln(e17) on your calculator. Your calculator probably
only has log base 10, log(x), and base e, ln(x).
Example
Q?
Solve for x, e2x−2 = 7
A.
Take the ln() of both sides.
e2x−2 = 7
ln(e2x−2) = ln(7)
2x − 2 = ln(7)
2x = ln(7) + 2
ln(7) + 2
x =
2
Note that ln(7) can be approximated on a
calculator, it’s just a decimal number and is treated as
such in the above algebra.
Example
Q?
Solve for x, ln(x − 7) = 29
Example
1
Q?
Find the domain of the function y = ln(x−2)
A.
There is no root sign. In order not to divide
by zero,
ln(x − 2) 6= 0
eln(x−2) 6= e0
x − 2 6= 1
x 6= 3
Now we just have to worry about only taking the ln()
of a strictly positive number, that is, x − 2 > 0, x > 2.
Therefore the domain of the function is (2, 3) ∪ (3, ∞).
2.2 Change of Base Formula
Ok, if my calculator only has log base 10 and base e,
how do I evaluate something like log3(4)? We change
it into base 10 or base e.
loga(x)
loga(b)
ln(x) log(x)
logb(x) =
=
ln(b)
log(b)
logb(x) =
Because we get to choose a in the above formula
we can make a = e or a = 10 to get the second
equations. So log3(4) =
ln(4)
ln(3)
' 1.261859 .
2.3 The Logarithm Rules
logb(AC)
A
logb( )
C
logb(A)x
1
logb( )
C
= logb(A) + logb(C)
= logb(A) − logb(C)
= x logb(A)
= − logb(C)
All the above rules are valid for ln(x) also.
We can use them to solve for x and to simplify
equations. Before we do some examples, let me
say what these rule do not say. It is not true that
logb (A)
logb(A − C) = log
or logb(A + C) = logb(A) logb(C) .
b (C)
These two misconceptions are common. Don’t fall for
it.
Example
Q?
Solve for x, ln(x) + ln(x + 1) = ln(2)
A rule of thumb for solving for x is: If the x is
inside a ln() use the exponential function to get it out,
if the x is inside an exponential function, take the ln()
to get it out.
Section 1.6, # 33, 35-40, 42, 47, 49-52.
Submit 38, 42, 50 Friday.