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Problem 1.1.12
Find the equation of the line: ( 12 , 1) and (1, 2) on line.
Solution: The slope is
or y = 2x.
2−1
1− 12
= 2. The equation of the line is y − 2 = 2(x − 1)
Problem 1.1.24
Find the equation of the line: parallel to y − x = 13; y-intercept is − 12 .
Solution: The slope is 1. So y = x − 12 .
Problem 1.1.36
How many units must you move in the y-direction is you start on a line of
slope .2 and move 5 units in the x-direction?
Solution: You must move (.2)(5) = 1 unit in the y direction.
Problem 1.1.52
A restaurant paying 5.75 had a quit ratio of .2. When the pay was raised to
6, the quit ratio was .18.
a) Assuming a linear relationship between wagex and Q(x) the quit ratio,
find Q(x).
= −.08.
Solution: The change in quit ratio over the change in wage is − .02
.25
So Q(x) − .18 = −.08(x − 6) or Q(x) = −.08x + .66.
b) When is Q(x) = .1?
Solution Q(x) = −0.8x + .66 = .1 ⇒ −0.8x = −.56. So x = 7.
Problem 1.1.68
Prove Slope Property 5 of straight lines.
Solution: Consider the picture in the book. We have two perpendicular
lines l1 and l2 intersecting at the origin with slopes m1 > m2 respectively.
By slope property 1, we know that along the x = 1 line, the distance from
1
l1 to the x-axis is m1 and the distance from l2 to the x-axis is m2 . So the
picture is correct.
The triangle bounded by the x-axis, the x = 1 line, and l1 is right, and
so by Pythagoras’ Theorem a2 = 1 + m21 . Similarly, we have b2 = 1 + m22
for the bottom right triangle. The last right triangle is the one formed by
x = 1, l1 and l2 , and the Pythagorean relationship is a2 + b2 = (m1 − m2 )2 .
Substituting into the last equation the other two equations, we have
1 + m21 + 1 + m22 = (m1 − m2 )2 = m21 − 2m1 m2 + m22
Simplifying, we get m1 m2 = −1.
Problem 1.2.22
Find the slope of the tangent line to the curve y = x2 at the point where
x = − 12 .
Solution: The slope at x is 2x, thus the slope of the tangent line at x = − 12
is -1.
Problem 1.2.28
Find the point on the graph of y = x2 where the tangent line is parallel to
the line 3x − y = 2.
Solution: The slope of that line is 3, so we must find the point on the curve
whose slope is 3 as well. Since the slope at x of the curve x2 is 2x, thus at
( 32 , 49 ) the tangent line is parallel to the given line.
Problem 1.2.30
Find the slope of the curve y = x3 at the point ( 23 , 27
).
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Solution: We simply plug
3
2
into 3x2 and get
27
4
as our answer.
Problem 1.2.36
Find a, f (a), and the slope of the curve at (a, f (a)).
Solution: (a, f (a)) lies at the intersection of the two lines y = 2 − x and
2
y = − 15 x + 65 . Substituting for y, we have the following equation
1
6
2−x=− x+
5
5
Collecting like terms, we deduce that x = 1. Thus a = 1. Now f (a) =
2 − a = 1. The slope of the curve at that point is simply the slope of the
tangent line, which is the line y = 2 − x. Since this line has slope −1, we
know that the slope of the curve at (a, f (a)) is -1 as well.
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Problem Use the Power Rule to find the derivative of (x− 5 )2 .
3
6
Solution: Using the exponent rule we know that (x− 5 )2 = x− 5 . Now ap11
plying the Power Rule, the derivative is − 56 x− 5 .
Problem Let L be the tangent line to the curve f (x) = x4 − 6 at (2, 10).
Let P be the line perpendicular to L and also passing through (2, 10). Find
the equation for P .
Solution: An application of the Power Rule gives us f 0 (x) = 4x3 . Thus at 2,
the derivative is 32. So the slope of L is 32. This implies that the slope of
1
1
1
1
. Thus the equation for P is y −10 = − 32
(x−2) or y = − 32
x+10 16
.
P is − 32
Problem 1.3.6
Find The derivative of f (x) =
Solution:
√1
x
√1 .
x
1
3
= x− 2 . So by the Power Rule f 0 (x) = − 12 x− 2 = − 2x1√x .
Problem 1.3.30
If f (x) = x12 , find f (5) and f 0 (5).
Solution: Using the Power Rule, we get that f 0 (x) =
−2
and f 0 (5) = 125
.
−2
.
x3
So f (5) =
1
25
Problem 1.3.44
The line y = ax+b is tangent to y = x3 at the point (−3, −27). Find a and b.
Solution: The derivative of x3 is 3x2 . Thus at x = −3 the slope of the
tangent line is 27. So the equation of the tangent line is y + 27 = 27(x + 3)
3
or y = 27x + 54. Therefore a = 27 and b = 54.
Problem 1.3.54
dy
if y = x−4
dx
Solution:
dy
dx
= −4x−5 .
Problem 1.3.62
Estimate f 0 (1).
Solution: Evaluating the slopes of the chords from (1, .8) to the the other
points, we get the following values:
1.3 − .8
1.1 − .8
1.4 − .8
= 1,
= 1.25,
= 1.5
1.6 − 1
1.4 − 1
1.2 − 1
This sequence of numbers is increasing, and one would guess that f 0 (1) > 1.5,
say for example 1.75.
Problem 1.3.68
√
Compute the difference quotient when f (x) = x3 − 2x2 + 5.
Solution:
f (x + h) − f (x)
(x + h)3 − 2(x + h)2 +
=
h
h
=
√
5 − x3 + 2x2 −
√
5
x3 + 3x2 h + 3xh2 + 3h3 − 2x2 − 4xh − 2h2 − x3 + 2x2
h
2
= 3x + 3xh + 3h2 − 4x − 2h
Problem 1.3.76
Apply the three step method to compute the derivative of f (x) = 2x2 + x.
Solution:
Step 1.
f (x + h) − f (x)
2(x + h)2 + x + h − 2x2 − x
=
h
h
4
=
2x2 + 4xh + 2h2 + x + h − 2x2 − x
= 4x + 2h + 1
h
Step 2.
Letting h → 0, this expression becomes 4x + 1.
Step 3.
Thus the derivative f 0 (x) = 4x + 1.
Problem 1.3.88
Show that
d
d
f (x) =
(f (x) + c)
dx
dx
Solution: If one were to solve it pictorially, simply draw a differentiable
function f , and then pick a c and draw out f + c on the same plane. Then
select a point on f and draw the tangent line. Then select the point on f + c
directly above or below the previous point and draw its tangent line as well.
You will notice that these two lines are parallel, telling us that f and f + c
have the same slope at all x.
To solve it algebraically, simply note the following:
f (x + h) + c − (f (x) + c)
f (x + h) − f (x)
=
h
h
Thus their difference quotients are the same, and so therefore their derivatives
are the same as well.
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