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Problem of the Day
Let f and g be odd functions. If p, r, and s are
nonzero functions defined as follows, which must
be odd?
I. p(x) = f(g(x))
II. r(x) = f(x) + g(x)
III. s(x) = f(x)g(x)
A) I
B) II
C) I and II
D) II and III
E) I, II, and III
Problem of the Day
Let f and g be odd functions. If p, r, and s are
nonzero functions defined as follows, which must
be odd?
I. p(x) = f(g(x))
II. r(x) = f(x) + g(x)
III. s(x) = f(x)g(x)
A) I
B) II
C) I and II
D) II and III
To be odd f(-x) = f(x)
E) I, II, and III
velocity
A person is holding a
grapefruit 6 feet above
the ground and tosses
it into the air.
height
6
time
time
feet
0
6
1 2
3
4
5 6
90 142 162 150 106 30
How far does the grapefruit travel in the first
second?
2nd second?
velocity
A person is holding a
grapefruit 6 feet above
the ground and tosses
it into the air.
height
6
time
time
feet
0
6
1 2
3
4
5 6
90 142 162 150 106 30
How far does the grapefruit travel in the first
second?
2nd
52
84 second?
velocity
A person is holding a
grapefruit 6 feet above
the ground and tosses
it into the air.
height
6
time
time
feet
0
6
1 2
3
4
5 6
90 142 162 150 106 30
Find the average velocity over the interval 4< t <
5. What is the significance of this answer?
velocity
A person is holding a
grapefruit 6 feet above
the ground and tosses
it into the air.
height
6
time
time
feet
0
6
1 2
3
4
5 6
90 142 162 150 106 30
Find the average velocity over the interval 4< t < 5. What is
the significance of this answer?
Δy = y(5) - y(4) = 106 - 150 = -44 ft/sec
Δx
5-4
1
The grapefruit is falling at a rate of 44 ft per sec
velocity
A person is holding a
grapefruit 6 feet above
the ground and tosses
it into the air.
height
6
time
time
feet
0
6
1 2
3
4
5 6
90 142 162 150 106 30
Find the average velocity over the interval 1< t <
4. What is the significance of this answer?
velocity
A person is holding a
grapefruit 6 feet above
the ground and tosses
it into the air.
height
6
time
time
feet
0
6
1 2
3
4
5 6
90 142 162 150 106 30
Find the average velocity over the interval 1< t < 4. What is the
significance of this answer?
Δy = y(4) - y(1) = 150 - 90 = 20 ft/sec
Δx
4-1
3
The grapefruit is rising at a rate of 20 ft per sec
velocity
(slope of secant line gives average
velocity)
(4, 150)
height
(1, 90)
(5, 106)
6
time
Δy = y(4) - y(1) = 150 - 90 = 20 ft/sec
Δx
4-1
3
(slope of secant line gives average
velocity)
velocity
(x + Δx, f(x + Δx))
(x, f(x))
x + Δx
{
x
Δx
Δy = f(x + Δx) - f(x)
Δx
(x + Δx) - x
velocity
(x + Δx, f(x + Δx))
(to find instantaneous velocity we want
Δx to be as small as possible or for Δx to
approach 0)
(x, f(x))
x + Δx
{
x
Δx
lim f(x + Δx) - f(x)
Δx 0
(x + Δx) - x
demo
(to find instantaneous velocity we want
Δx to be as small as possible or for Δx to
approach 0)
velocity
(x + Δx, f(x + Δx))
(x, f(x))
x + Δx
{
x
Δx
Our secant line becomes a tangent line
lim f(x + Δx) - f(x)
Δx 0
(x + Δx) - x
The limit used to define the slope of a tangent line
is also used to define one of the two fundamental
operations of calculus - differentiation.
The derivative of f at x is given by:
f ' (x) = lim
Δx 0
f(x + Δx) - f(x)
Δx
(provided the limit exists)
Notation for derivatives
f ' (x)
dy
dx
y'
d[f(x)]
dx
Each represents
the derivative of y with respect to x
Dx[y]
Find the derivative of
3
f(x) = x + 6x
f ' (x) = lim
Δx
f(x + Δx) - f(x)
0
Δx
(x + Δx)3 + 6(x + Δx) - (x3 + 6x)
Δx
x3 + 3x2Δx + 3x(Δx)2 + (Δx)3 + 6x + 6Δx - x3- 6x
Δx
Find the derivative of
3
f(x) = x + 6x
x + 3x Δx + 3x(Δx) + (Δx) + 6x + 6Δx - x - 6x
3
2
2
3
3
Δx
3x2Δx + 3x(Δx)2 + (Δx)3 + 6Δx
Δx
Δx(3x2 + 3xΔx + (Δx)2 + 6
Δx
Find the derivative of
3
f(x) = x + 6x
Δx(3x2 + 3xΔx + (Δx)2 + 6
Δx
3x2 + 3xΔx + (Δx)2 + 6
2
f ' (x) = 3x + 6
Find the derivative of
2
f(x) = x + 4
Find the derivative of
2
f(x) = x + 4
f ' (x) = 2x
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