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Tutorial 6 - April 04, 2014
Questions related to SECTION 4.3
1. Sketch the graph of f (x) = x4 − 6x2
(a) Domain: (−∞, ∞)
(b) x− and y− intercepts:
x−intercept: y = 0
⇔
x4 − 6x2 = 0
So the x−intercepts are (0, 0),
y−intercept: x = 0
⇔
√
(− 6, 0),
x = 0,
√
x = − 6,
x=
√
6
√
( 6, 0)
⇔y=0
So the y−intercept is (0, 0).
(c) Critical Points and Local Extreme Values:
i. Find f 0 (x)
f 0 (x) = 4x3 − 12x = 4x(x2 − 3)
ii. Find all critical points:
f 0 (x) = 0 : f 0 (x) = 4x(x2 − 3) = 0
⇔
x = 0,
√
x = − 3,
x=
√
3
f 0 (x) is not defined: f 0 (x) is defined for all real x.
iii. Increasing and Decreasing Intervals:
√
√
x
x<
− 3
<x<
0
<x<
3
<x
0
f (x) −
0
+
0
−
0
+
f (x) & loc. min
%
loc. max
&
loc. min %
iv. Evaluate f (x) at each critical point:
√
√
local minimums at (− 3, −9) , ( 3, −9) and local maximum at (0, 0)
(d) Determine Concavity and Points of Inflection:
00
i. Find f (x)
00
f (x) = 12x2 − 12 = 12(x2 − 1) = 12(x − 1)(x + 1)
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Department of Mathematics, Eastern Mediterranean University
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Tutorial 6 - April 04, 2014
ii. Find all potential points of inflection:
00
f (x) = 0 :
00
f (x) = 12(x − 1)(x + 1) = 0
00
⇔
x = 1,
x = −1
00
f (x) is not defined: f (x) is defined for all real x.
iii. Intervals of Concavity:
x
x<
00
f (x) +
f (x)
∪
−1
<x<
1
<x
0
−
0
−
infl.
∩
infl.
∪
point
point
iv. The points of inflection are : (−1, −5) and (1, −5)
(e) Asymptotes:
Horizontal Asymptote: None
Vertical Asymptote: None
Oblique Asymptote: None
(f) Sketch the graph:
10
5
-4
-2
2
4
-5
-10
(g) The range of f (x) is [−9, ∞).
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Department of Mathematics, Eastern Mediterranean University
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Tutorial 6 - April 04, 2014
2. Sketch the graph of y =
(x + 1)2
1 + x2
(a) Domain: x ∈ R
(b) x− and y− intercepts:
x−intercept: y = 0
⇔
(x + 1)2
=0
1 + x2
⇔ (x + 1)2 = 0
⇔ x = −1
So the x−intercept is (−1, 0).
y−intercept: x = 0
⇔y=
(0 + 1)2
1 + 02
⇔y=1
So the y−intercept is (0, 1).
(c) Critical Points and Local Extreme Values:
i. Find f 0 (x)
f 0 (x) =
(1 + x2 ) · 2(x + 1) − (x + 1)2 · 2x
2(1 − x2 )
=
(1 + x2 )2
(1 + x2 )2
ii. Find all critical points:
f 0 (x) = 0 :
f 0 (x) =
2(1 − x2 )
=0
(1 + x2 )2
⇔
x = −1 and x = 1
f 0 (x) is not defined: f 0 (x) is defined for all real x.
iii. Increasing and Decreasing Intervals:
x
x<
−1
<x<
1
<x
0
f (x) −
0
+
0
−
f (x) & loc. min.
%
loc. max. &
iv. Evaluate f (x) at each critical point:
local minimum at (−1, 0) and local maximum at (1, 2)
(d) Determine Concavity and Points of Inflection:
00
i. Find f (x)
00
f (x) =
(1 + x2 )2 · 2(−2x) − 2(1 − x2 )[2(1 + x2 ) · 2x]
4x(x2 − 3)
=
(1 + x2 )4
(1 + x2 )3
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Department of Mathematics, Eastern Mediterranean University
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Tutorial 6 - April 04, 2014
ii. Find all potential points of inflection:
00
f (x) = 0 :
00
f (x) =
00
4x(x2 − 3)
=0
(1 + x2 )3
√
⇔ x = 0, x = ± 3
00
f (x) is not defined: f (x) is defined for all real x.
iii. Intervals of Concavity:
√
x
x< − 3 <x<
0
<x<
00
f (x) −
0
+
0
−
f (x)
∩
infl.
∪
infl.
∩
point
point
√
3
<x
0
+
infl.
∪
point
√
√
iv. The points of inflection are : (− 3, 0.1340) , (0, 1) and ( 3, 1.8660)
(e) Asymptotes:
Horizontal Asymptote:
x2 1 + x2 + x12
x2 + 2x + 1
(x + 1)2
=1
= lim
= lim
lim
x→±∞
x→±∞
x→±∞ 1 + x2
1 + x2
x2 1 + x12
So we have a horizontal asymptote at y = 1.
Vertical Asymptote: None
Oblique Asymptote: None
(f) Sketch the graph:
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Department of Mathematics, Eastern Mediterranean University
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Tutorial 6 - April 04, 2014
2
1
-6 -5 -4 -3 -2 -1 0
1
2
3
4
5
6
-1
(g) The range of f (x) is [0, 2].
Questions related to SECTION 4.5
1. (a) Write the equation of the line that represents the linear approximation to
the following functions at the given point a.
(b) Use the linear approximation to estimate the given function value.
(c) Compute the percent error in your approximation by the formula:
100 ·
i. f (x) = 12 − x2 ;
a = 2;
|approx − exact|
|exact|
f (2.1)
(a) f (a) = f (2) = 8 and f 0 (a) = −2a = −4
so the linear approximation has the equation:
y = L(x) = f (a) + f 0 (a)(x − a) = 8 − 4(x − 2) = −4x + 16
(b) f (2.1) ≈ L(2.1) = 7.6
(c) The percent error is :
100 ·
|7.6 − 7.59|
≈ 0.13%
|7.59|
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Department of Mathematics, Eastern Mediterranean University
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Tutorial 6 - April 04, 2014
ii. f (x) = sin x;
a = π/4;
f (0.75)
√
(a) f (a) = f (π/4) = sin(π/4) =
2
2
and f 0 (a) = cos a = cos(π/4) =
√
2
2
so the linear approximation has the equation:
π
π
π
y = L(x) = f (a) + f 0 (a)(x − a) = sin + cos
x−
=
4
4
4
√
√ √ π
π
2
2
2
=
+
x−
=
x+1−
2
2
4
2
4
(b) f (0.75) ≈ L(0.75) ≈ 0.68
(c) The percent error is :
iii. f (x) = ln (1 + x);
100 ·
a = 0;
|0.68 − sin 0.75|
≈ 0.064%
| sin 0.75|
f (0.9)
(a) f (a) = f (0) = ln 1 = 0 and f 0 (a) =
1
1
=
=1
1+a
1+0
so the linear approximation has the equation:
y = L(x) = f (a) + f 0 (a)(x − a) = 0 + 1(x − 0) = x
(b) f (0.9) ≈ L(0.9) = 0.9
(c) The percent error is :
iv. f (x) = ex ;
a = 0;
100 ·
|0.9 − ln 1.9|
≈ 40%
| ln 1.9|
f (0.05)
(a) f (a) = f (0) = e0 = 1 and f 0 (a) = ea = e0 = 1
so the linear approximation has the equation:
y = L(x) = f (a) + f 0 (a)(x − a) = 1 + 1(x − 0) = 1 + x
(b) f (0.05) ≈ L(0.05) = 1.05
(c) The percent error is :
100 ·
|1.05 − e0.05 |
≈ 0.12%
|e0.05 |
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Department of Mathematics, Eastern Mediterranean University
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Tutorial 6 - April 04, 2014
2. Use linear approximations to estimate the following quantities. Choose a value
of ’a’ to produce a small error.
√
(a) 146
Let f (x) =
f 0 (x) =
√
x
1
√
2 x
and
⇒
a = 144. Then
f 0 (a) =
√1
2 144
=
f (a) =
√
144 = 12
1
24
So the linear approximation near a = 144 is
L(x) = f (a) + f 0 (a)(x − a) = 12 +
1
(x
24
− 144)
Therefore
√
(b)
√
3
146 = f (146) ≈ L(146) = 12 +
1
1
(146 − 144) = 12
24
12
65
1
and
Let f (x) = x 3
f 0 (x) = 13 x
−2
3
f 0 (a) = 13 64
⇒
1
a = 64. Then
−2
3
f (a) = 64 3 = 4
1
48
=
So the linear approximation near a = 64 is
L(x) = f (a) + f 0 (a)(x − a) = 4 +
1
(x
48
− 64)
Therefore
√
3
65 = f (65) ≈ L(65) = 4 +
1
1
(65 − 64) = 4
48
48
(c) e0.06
Let f (x) = ex
f 0 (x) = ex
⇒
and
a = 0. Then
f (a) = e0 = 1
f 0 (a) = e0 = 1
So the linear approximation near a = 0 is
L(x) = f (a) + f 0 (a)(x − a) = 1 + 1(x − 0)
Therefore
e0.06 = f (0.06) ≈ L(0.06) = 1 + 1(0.06 − 0) = 1.06
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Department of Mathematics, Eastern Mediterranean University
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Tutorial 6 - April 04, 2014
(d) cos 31◦
Let f (x) = cos x
f 0 (x) = − sin x
a = π6 (= 30). Then f (a) =
and
⇒
f 0 (a) = − sin 30 =
√
3
2
−1
2
So the linear approximation near a = 30◦ is :
L(x) = f (a) + f 0 (a)(x − a) =
√
3
2
− 12 (x − π6 )
Therefore
◦
cos 31 = cos
31π
180
=f
31π
180
≈L
31π
180
√
3 1
=
−
2 2
31π π
−
180
6
≈ 0.857
Note that we must convert 31◦ to radians before applying the linear approximation formula.
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Department of Mathematics, Eastern Mediterranean University
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