MATH 147
Lab Key #4
2/23/2016
(1) Use the Sandwich Theorem to evaluate lim e4x sin x.
x →−∞
Solution: Note that for all x ∈ R,−1 ≤ sin x ≤ 1. Therefore,−e4x ≤ e4x sin x ≤
e4x . Now, note that lim −e4x = 0 and lim e4x = 0. Hence by the Sandwich
x →−∞
Theorem,
x →−∞
lim e4x sin x = 0.
x →−∞
(2) Evaluate each limit:
sin (8x )
(a) lim
2x
x →0
3 − 3 cos (2x )
(b) lim
8x
x →0
Solution: (a)
lim
x →0
sin (8x )
4 sin (8x )
sin (8x )
= lim ·
= 4 · lim
= 4 · 1 = 4.
2x
2x
8x
x →0 4
x →0
(b)
3 [1 − cos (2x )]
3
1 − cos (2x )
3
3 − 3 cos (2x )
= lim
= · lim
= · 0 = 0.
8x
4 · 2x
4 x →0
2x
4
x →0
x →0
lim
(3) Use the Intermediate Value Theorem to show that there is a solution of the
equation x3 − 2x + 3 = 0 in the interval (−3, −1) .
Solution: Let f ( x ) = x3 − 2x + 3. Then f (−3) = (−3)3 − 2 (−3) + 3 =
−27 + 6 + 3 = −18, and f (−1) = (−1)3 − 2 (−1) + 3 = −1 + 2 + 3 = 4.
Hence, f (−3) < 0 < f (−1) . Therefore, since f is a continuous function, by the
Intermediate Value Theorem, there exists a c ∈ (−3, −1) such that f (c) = 0.
This c is a solution to the given equation.
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(4) Find an equation of the tangent line to the curve y = 3x2 − 4x + 7 at x = 2.
Solution: To find the equation of a line, we need its slope and a point on it. If
f ( x ) = 3x2 − 4x + 7, then a point on the line is (2, f (2)) .
f (2) = 3 · 4 − 4 · 2 + 7 = 12 − 8 + 7 = 11.
Hence, a point on the line is (2, 11) . Furthermore, the slope is given by the
f 0 (2) .
f 0 ( x ) = 6x − 4.
∴ f 0 (2) = 6 · 2 − 4 = 12 − 4 = 8.
Therefore, m = 8. Now, using point-slope form, we get the equation of our
tangent line as
y − 11 = 8 ( x − 2) .
∴ y − 11 = 8x − 16.
∴ y = 8x − 5.
Hence, the equation of the tangent line to f ( x ) at x = 2 is given by y = 8x − 5.
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(5) Differentiate eachfunction.
(a) f ( x ) = 6x2 + 2x 4x3 + 2x2 − x
x2 − 3x
.
(b) g ( x ) = 2
x +1
Solution: (a) We can use the product rule to differentiate f :
i
d h
i
d h 2
6x + 2x 4x3 + 2x2 − x + 6x2 + 2x ·
4x3 + 2x2 − x .
dx
dx
∴ f 0 ( x ) = (12x + 2) 4x3 + 2x2 − x + 6x2 + 2x 12x2 + 4x − 1 .
f 0 (x) =
∴ f 0 ( x ) = 48x4 + 24x3 − 12x2 + 8x3 + 4x2 − 2x + 72x4 + 24x3 − 6x2 + 24x3 + 8x2 − 2x.
∴ f 0 ( x ) = 120x4 + 80x3 − 6x2 − 4x.
h
i
∴ f 0 ( x ) = 2x 60x3 + 40x2 − 3x − 2 = 2x 20x2 (3x + 2) − (3x + 2) .
∴ f 0 ( x ) = 2x (3x + 2) 20x2 − 1 .
(b) We can use the quotient rule to differentiate g :
0
g (x) =
d
dx
x2 − 3x
∴ g0 ( x ) =
x2 + 1 − x2 − 3x
d
dx
x2 + 1
2
( x 2 + 1)
(2x − 3) x2 + 1 − x2 − 3x (2x )
( x 2 + 1)
∴ g0 ( x ) =
2
2x3 − 3x2 + 2x − 3 − 2x3 + 6x2
∴ g0 ( x ) =
( x 2 + 1)
2
3x2 + 2x − 3
( x 2 + 1)
2
.
.
.
.
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