Download Solution for Midterm II

MATH 113/114 Midterm 2 Solution
Question 1. (5 points) Find the derivative of the following functions
(DO NOT SIMPLIFY):
√
(a) f (x) = (x2 + 3)4 · 3 x3 + 2x
(b) f (x) = cos(cos(cos2 x))
Solution
(a)
f 0 (x) = 4(x2 + 3)3 · (2x) ·
p
3
2
1
x3 + 2x + (x2 + 3)4 · (x3 + 2x)− 3 · (3x2 + 2)
3
(b)
f 0 (x) = − sin(cos(cos2 x)) · (− sin(cos2 x)) · 2 cos x · (− sin x)
Question 2. (3 points) Use implicit differentiation to find an equation of the tangent line to the curve
x3 + y 3 = 2xy
at the point (1, 1).
2
dy
dy
2y−3x
dy
= 2y + 2x dx
, and thus dx
= 3y
Differentiation gives 3x2 + 3y 2 dx
2 −2x . Hence, the slope of the
dy
tangent line at point (1, 1) is m = dx
= 2−3
3−2 = −1. Therefore, the equation of the tangent at (1, 1) is
Solution
(1,1)
y − 1 = −(x − 1), namely, y = −x + 2.
Question 3. (3 points) Evaluate the limit lim
sin(x−3)
2
x→3 x −4x+3
Solution
lim
x→3
sin(x − 3)
sin(x − 3)
1
= lim
=
x2 − 4x + 3 x→3 (x − 3)(x − 1)
2
Question 4. (3 points) The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the
triangle is increasing at a rate of 20 cm2 /min. At what rate is the base of the triangle changing when the
altitude is 10 cm and the area is 100 cm2 .
Solution Since the area of the triangle A = 100 and the height of the triangle h = 10, by the formula
1 dh
1 db
A = 21 hb, we obtain the base of the triangle b = 20. Differentiations gives dA
dt = 2 dt b + 2 h dt . Plugging into
dA
dh
db
dt = 20, dt = 1, b = 20 and h = 10, we obtain dt = 2, namely, the base of the triangle is increasing at a
rate of 2 cm/min.
Question 5. (3 points) Use linear approximations to estimate the number (1.01)5 .
Solution
Let f (x) = x5 and a = 1. We have f (1) = 1 and f 0 (x) = 5x4 , and thus f 0 (1) = 5. Hence
the linear approximations to f (x) near x = 1 is L(x) = f (1) + f 0 (1)(x − 1) = 1 + 5(x − 1). Therefore,
(1.01)5 ≈ 1 + 5(1.01 − 1) = 1.05.
Question 6. (3 points) Find the absolute maximum and minimum values of the function f (x) =
x
on
x2 + 1
the interval [0, 3].
Solution
Since
f 0 (x) =
x2 + 1 − x · 2x
1 − x2
(1 + x)(1 − x)
=
=
,
(x2 + 1)2
(x2 + 1)2
(x2 + 1)2
the critical numbers are 1 and −1. Note that −1 does not lie in the interval [0, 3]. Comparing f (0) =
3
0, f (1) = 12 , f (3) = 10
, we conclude that f (x) has absolute maximum 12 when x = 1 and absolute minimum
0 when x = 0.
√
Question 7. (3 points) Verify that the function f (x) = x x + 5 satisfies the hypotheses of Roll’s Theorem
on the interval [−5, 0]. Then find all numbers c that satisfy the conclusion of Roll’s Theorem.
1
Solution
Note that
f 0 (x) =
√
2x + 10 + x
3x + 10
1
√
=
= √
.
x+5+x √
2 x+5
2 x+5
2 x+5
Hence f 0 (x) exists for all value (−5, +∞). This implies that f (x) is continuous on [−5, 0] and differentiable
on (−5, 0). We also have f (−5) = f (0) = 0. Therefore, by Roll’s Theorem, there exists c ∈ (−5, 0) such that
10
√
f 0 (c) = 0. On the other hand, let f 0 (x) = 23x+10
= 0. Solving it gives x = − 10
3 . Therefore, c = − 3 is the
x+5
number that satisfies the conclusion of Roll’s Theorem.
Question 8. (2 points) Evaluate the limit lim
x→1
Solution
x2010 −1
x−1
Let f (x) = x2010 . By definition of derivative of f (x) at x = 1, we have
f (x) − f (1)
x2010 − 1
= lim
.
x→1
x→1
x−1
x−1
f 0 (1) = lim
On the other hand, f 0 (x) = 2010x2009 , and thus f 0 (1) = 2010. Therefore,
x2010 − 1
= f 0 (1) = 2010.
x→1
x−1
lim
2