Download x whenever f(x)

If f passes the horizontal line test then it has an inverse,
f −1 , defined by the property f −1 (y) = x whenever f (x) = y
f (x) = 10x is increasing so it has an inverse for which we
use the special notation log y instead of f −1 .
x
y
100
2
x
y = 10
2
x
x=log(y)
100
y
2 ←→ 100
y = 10x expresses the relationship, writing y in terms of x.
x = log y expresses the SAME relationship, writing x in
terms of y. log y is the power to which we raise 10 to get y.
When composed with f , the inverse of f UNDOES the
action of f :
f −1 (f (x)) = x and f (f −1 (y) = y
log (10x ) = x and 10log y = y
The same thing applies when we use a different base, say
y = 5x with an inverse denoted by x = log5 y
f −1 (f (x)) = x and f (f −1 (y) = y
log5 (5x ) = x and 5log5 y = y
More examples:
log 1 = 0 because 100 = 1
log 1 = log 100 = 0
log5 125 = 3 because log5 125 = log5 53 = 3
ln (eA ) = A
Logarithm Rules (same apply for all ln x, log x, log3 x, etc)
I ln (AB) = ln A + ln B
A
II ln ( ) = ln A − ln B
B
III ln (AP ) = P ln A
To see why the first one is true, compare:
eln AB = AB
eln A+ln B = eln A eln B = AB
When e is raised to either power ln AB or ln A + ln B we
get the same result, namely AB. So they must be equal.
This is how ln y is defined: It is the power to which you
need to raise e in order to produce y.
Examples of problems solved using the Logarithm Rules
I ln (AB) = ln A + ln B
A
II ln ( ) = ln A − ln B
B
III ln (AP ) = P ln A
Problem 1. Solve 1000(3t/50 ) = 4000 for t
Divide by 1000 and then apply any logarithm function to
both sides (we choose ln or log because our calculators have
function keys for these and allow us to find approximate
values for them.
3t/50 = 4
ln (3t/50 ) = ln 4
t
ln 3 = ln 4
50
50 ln 4
t=
ln 3
Problem 2. Express log5 x in terms of ln
First write y = log5 x which means 5y = x. Then solve for y
using ln.
5y = x
ln 5y = ln x
y ln 5 = ln x
ln x
y=
ln 5
log5 x =
ln x
ln 5
Clicker Question:
log (
M −N
) =?
M +N
(a) 2 log M
(b) 2 log N
(c) −2 log N
(d) log (M − N ) − log (M + N ) (e) More than one of these
Clicker Question:
Without your calculator and using only properties of
logarithms, decide which of the following is the largest.
(a) log3 (30) − log3 (2)
(b) 2 log3 4
(c) log3 (3) + log3 (4)
(d)
log3 4
log3 2
Clicker Question:
Give a formula for the inverse of P = f (x) = 16e14x
1
(a) f (P ) = e−14x
16
ln 16
(c) f −1 (P ) = (
)P
14
−1
1
14
ln P
ln 16
P
1
d) f −1 (P ) =
ln ( )
14
16
(b) f
−1
(P ) =
y = cos t 0 ≤ t ≤ π; passes the horizontal line test so it
has an inverse, which we write as either arccos y or cos−1 y
(x,y)
t
π
t
This means t = arccos y is the angle t in [0, π] for which
cos t = y.
So arccos (0) = π/2 and arccos (1) = 0 and arccos (−1) = π
2
Find sin (arccos ( √ )).
5
2
Let t = arccos ( √ ) and draw a triangle that represents an
5
2
angle t with cos t = √ .
5
1/2
5
1
t
2
We can see that
1
2
1
sin t = √ , that is sin (arccos ( √ )) = √ .
5
5
5
2
1
π/6
1/2
3
arcsin (1/2) =
π
6
√
cos (arcsin (1/2)) =
3
2
Average velocity = displacement / time elapsed
A ball is dropped from 500 ft high. After t seconds it will
have traveled f (t) = t2 feet from where it was dropped.
What is the average velocity over the time t = 4 to t =
4+h?
f (4 + h) − f (4)
(4 + h)2 − (4)2
=
h
h
(16 + 8h = h2 ) − (16)
=
h
8h + h2
=
= 8 + h feet per second
h
Average velocity is 8 + h ft/sec over an elapsed time of h
seconds.
What is the instantaneous velocity at t = 4?
If f(t) is the position function the describes the motion of
an object moving along a straight line, then:
Average velocity over the time interval [a, a+h] is
f (a + h) − f (a)
h
Velocity is the limit of average velocities measured over
smaller and smaller time intervals.
Velocity at time t = a is
f (a + h) − f (a)
h→0
h
lim
(Also called the instantaneous velocity)
f (a + h) − f (a)
=L
h→0
h
means as h approaches the number 0 the value of
f (a + h) − f (a)
gets closer & closer to the number L.
h
lim
f (4 + h) − f (4)
= lim (8 + h) = 8
h→0
h→0
h
lim
y = f(x)
y
m sec
m tan
a
a+h
The slope of the secant line is msec =
x
f (a + h) − f (a)
h
f (a + h) − f (a)
h→0
h
The slope of the tangent line is mtan = lim
f (a + h) − f (a)
h
A ball is dropped from 500 ft high. After t seconds it will
have traveled f (t) = t2 feet from where it was dropped.
The slope of the tangent line is mtan = lim
h→0
The instantaneous velocity at t = 4 is
f (4 + h) − f (4)
= lim (8 + h) = 8
h→0
h→0
h
lim
which is exactly the same as the slope mtan of the tangent
line to the graph of (t) = t2 at the point (4,16).
Velocity and the slope of the tangent line are closely
connected. They are computed by taking the same limit.