Lab9

Math 125 Lab Section 63965
University of Kansas
February 14th, 2017
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Math 125 Lab Section 63965
February 14th, 2017
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Sections 3.1 and 3.2
Exercise 1. Find all a and b such that the function
2
ax + 4
if x < 1
f (x) =
3x 2 + bx if x ≥ 1
is differentiable for all x.
Answer: a = 7, b = 8
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February 14th, 2017
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Sections 3.2
Exercise 2. Find the points on the curve
y = (x − 1)2 (x + 2)
where the tangent line is horizontal.
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February 14th, 2017
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Sections 3.2
Exercise 2. Find the points on the curve
y = (x − 1)2 (x + 2)
where the tangent line is horizontal.
Solution: We need to find the points (x, y ) where y 0 (x) = 0. To compute
y 0 we could use product rule or multiply (x − 1)2 (x + 2) to get
y = x 3 − 3x + 2, anyway we get: y 0 = 3x 2 − 3. Solving 3x 2 − 3 = 0 we get
x = 1 and x = −1. Using these values in y = x 3 − 3x + 2 we get y = 0
and y = 4 respectively. Finally the desired points are: (1, 0) and (−1, 4)
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February 14th, 2017
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Sections 3.2
2
Exercise 3. For f (x) = 3x +5x−2
. Find the equations of the tangent and
x
1
normal line to f (x) when x = 1.
1
The normal line is the line that is perpendicular to the tangent line
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February 14th, 2017
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Sections 3.2
2
Exercise 3. For f (x) = 3x +5x−2
. Find the equations of the tangent and
x
1
normal line to f (x) when x = 1.
2
Solution: We know that the slope of the tangent line to f (x) = 3x +5x−2
x
at x = 1 is given by f 0 (1). To compute f 0 we can use quotient rule or do
the following:
3x 2 + 5x − 2
3x 2 5x
2
2
=
+
− = 3x + 5 − = 3x + 5 − 2x −1
x
x
x
x
x
0
And then take the derivative, anyway we get f (x) = 3 + 2x −2 and
f 0 (1) = 5, and thus mtangent = 5, mnormal = −1
5 . Finally since both tangent
and normal line to f (x) at x = 1 pass through (1, f (1)) = (1, 6), we get:
y − 6 = 5(x − 1)
−1
normal line to f (x) at x = 1: y − 6 =
(x − 1)
5
tangent line to f (x) at x = 1:
1
The normal line is the line that is perpendicular to the tangent line
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Math 125 Lab Section 63965
February 14th, 2017
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Section 3.2
* Exercise 4. Find the equation of the tangent line to f (x) = x 2 + 4 that
passes through the point (1, −4)
Answer:There are two lines: y − 20 = 8(x − 4) and y − 8 = −4(x + 2)
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February 14th, 2017
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Sections 3.3
[x n ]0 = nx n−1
[ax ]0 = ax ln(a)
[f (x)g (x)]0 = g (x)f 0 (x) + f (x)g 0 (x)
f (x)
g (x)
0
=
g (x)f 0 (x) − f (x)g 0 (x)
g 2 (x)
[f ( g (x) )]0 = f 0 ( g (x) ) · g 0 (x)
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February 14th, 2017
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Sections 3.3
Exercise 1.) You are given the following table
f (4)
1
f 0 (4)
2
g (4)
2
g 0 (4)
1
Compute
1
(fg )0 (4)
Answer: 1.)
0
f
and
(4)
g
(fg )0 (4)
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2
H 0 (4) where H(x) =
x 2 g (x)
f (x)
0
f
= 5 and
(4) = 34 . 2.) H 0 (4) = −32
g
Math 125 Lab Section 63965
February 14th, 2017
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Sections 3.3
Exercise 2.) Compute the derivative of the following functions
1
f (x) = (x 2 − 2x + 1)(2x − x)
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2
g (x) =
Math 125 Lab Section 63965
x − 3x 2
e x + 2x
February 14th, 2017
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Sections 3.3
Exercise 2.) Compute the derivative of the following functions
1
f (x) = (x 2 − 2x + 1)(2x − x)
2
g (x) =
x − 3x 2
e x + 2x
Solution
1
0
f 0 (x) = (x 2 − 2x + 1)(2x − x)
= (2x − x)(x 2 − 2x + 1)0 + (x 2 − 2x + 1)(2x − x)0
2
= (2x − x)(2x − 2) + (x 2 − 2x + 1)(2x ln(2) − 1)
0
x − 3x 2
0
g (x) =
e x + 2x
(e x + 2x)(x − 3x 2 )0 − (x − 3x 2 )(e x + 2x)0
=
(e x + 2x)2
(e x + 2x)(1 − 6x) − (x − 3x 2 )(e x + 2)
=
(e x + 2x)2
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February 14th, 2017
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Section 3.1, 3.2 and 3.3
A Each of the following limits represents a derivative f 0 (a). Find f (x)
and “a” and use this to compute the limit.
1
2
3
x 4 + 2x + 1
x→−1
x +1
h+3
2
−8
lim
h→0
h
x+1
−3
lim x−1
x→2 x − 2
lim
4
(h − 2)4 − 16
h→0
h
5
(x 3 + 2x)(x 2 + 1) − 6
x→1
x −1
6
lim
lim
lim
h→0
2
2+h
√
e h + 1+h
−1
h
The function f (x) and the point “a” are not unique, but limit value is
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Math 125 Lab Section 63965
February 14th, 2017
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Section 3.1, 3.2 and 3.3
A Each of the following limits represents a derivative f 0 (a). Find f (x)
and “a” and use this to compute the limit.
1
2
3
x 4 + 2x + 1
x→−1
x +1
h+3
2
−8
lim
h→0
h
x+1
−3
lim x−1
x→2 x − 2
lim
4
(h − 2)4 − 16
h→0
h
5
(x 3 + 2x)(x 2 + 1) − 6
x→1
x −1
6
lim
lim
lim
h→0
Answer:
1
2
3
2
2+h
√
e h + 1+h
−1
h
2
f (x) = x 4 + 2x, a = −1. L: −2
f (x) = 2x , a = 3. L: 8 ln(2)
x +1
f (x) =
, a = 2. L: −2
x −1
4
5
6
f (x) = x 4 , a = −2. L: −32
f (x) = (x 3 + 2x)(x 2 + 1), a = 1.
L: 16
x+1√
f (x) = e x−1
, a = 1. L: −1/4
+ x
The function f (x) and the point “a” are not unique, but limit value is
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Math 125 Lab Section 63965
February 14th, 2017
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