MATH 100, HOMEWORK 2 SOLUTIONS
Section 2.7, #22:
g(x) − g(0)
1 − x3 − 1
−x3
= lim
= lim
= lim (−x2 ) = 0.
x→0
x→0
x→0 x
x→0
x−0
x
g 0 (0) = lim
We also have g(0) = 1. Therefore the equation of the tangent is y = 1 + 0 · x,
i.e. y = 1.
Section 2.8, #26:
1 3 + x + h
3+x −
h→0 h 1 − 3x − 3h
1 − 3x
1 (3 + x + h)(1 − 3x) − (3 + x)(1 − 3x − 3h)
= lim
h→0 h
(1 − 3x − 3h)(1 − 3x)
f 0 (x) = lim
1 (3 + x + h − 9x − 3x2 − 3xh) − (3 + x − 9x − 3x2 − 9h − 3xh)
h→0 h
(1 − 3x − 3h)(1 − 3x)
= lim
1
10h
h→0 h (1 − 3x − 3h)(1 − 3x)
= lim
= lim
h→0
10
10
=
.
(1 − 3x − 3h)(1 − 3x)
(1 − 3x)2
The domain of both f and f 0 is the set of all x where 1 − 3x 6= 0, ie.
(−∞, 1/3) ∪ (1/3, ∞).
Section 2.8, #28:
√
√
1 1
1 1 t− t+h
√
√√
g (t) = lim
− √ = lim
h→0 h
h→0 h
t
t+h
t t+h
√
√
√
√
1 ( t − t + h)( t + t + h)
√
√√
√
= lim
h→0 h
t t + h( t + t + h)
0
t−t−h
√
√
= lim √ √
h→0 h t t + h( t +
t + h)
−1
−1
1
√
√ = − t−3/2 .
√
=
= lim √ √
h→0
2
t·2 t
t t + h( t + t + h)
1
The domain of both g and g 0 is the set of all t 6= 0, ie. (−∞, 0) ∪ (0, ∞).
Section 3.1, #30:
v=
√
1 2
x+ √
= (x1/2 + x−1/3 )2 = x + 2x1/6 + x−2/3 ,
3
x
dv
1
2
1
2
= 1 + 2 · x−5/6 − x−5/3 = 1 + x−5/6 − x−5/3 .
dx
6
3
3
3
Section 3.1, #54: The line y = 1 + 3x has slope
we are looking
√ 03, hence
0
3/2 0
for tangent lines with slope 3. We have y = (x x) = (x ) = 23 x1/2 . We
first find all x where y 0 = 3:
3 1/2
x = 2, x1/2 = 2, x = 4.
2
At x = 4, y = 4 · 2 = 8. The tangent line that we are looking for has slope 3
and goes through the point (4, 8), therefore it has the equation
y = 8 + 3(x − 4), or equivalently y = 3x − 4.
2