MATH 135 (Practice) Midterm II(Fall 2015)
Instructor: Xiaowei Wang
Nov. 9th, 1:00pm-3:15pm, 2015
Problem 1. Evaluate the following limit
sin 7x
.(= 0)
x→0 cos 2x
1. lim
tan 3x
3 tan 3x
2x
3
.(= lim
·
= )
x→0 tan 2x
x→0 2 3x
tan 2x
2
2. lim
tan 7x
.(= 7)
x→0
x
3. lim
ex − 1
dex 4. lim
.(=
= 1)
x→0
x
dx x=0
Problem 2. Evaluate the derivative f 0 (x) of the following functions
1. sin x · ln x2 .
solution. f 0 (x) = cos x · ln x2 +
2 sin x
.
x
2. csc−1 (tan ex ).
solution. f 0 (x) =
3. y =
| tan ex |
−1
p
· sec2 ex · ex .
(tan ex )2 − 1
(x + 1)3/2 (x − 4)5/2
.
(5x + 3)2/3
solution. Using logarithm differentiation. We first notice that
ln f (x) =
3
5
2
ln(x + 1) + ln(x − 4) − ln(5x + 3)
2
2
3
Differentiate both side we obtain that
3
5
10
0
f (x) = f (x) ·
+
−
2(x + 1) 2(x − 4) 3(5x + 3)
(x + 1)3/2 (x − 4)5/2
3
5
10
=
·
+
−
2(x + 1) 2(x − 4) 3(5x + 3)
(5x + 3)2/3
Problem 3. Find
dy
d2 y
and
.
dx
dx2
1
x + y = sin y.
solution. Differentiating both sides of the above equation, we obtain 1 + y 0 = cos y · y 0 .
Hence
1
y0 =
(1)
cos y − 1
Next differentiating (1), we have
y 00 =
sin y
sin y · y 0
=
2
(cos y − 1)
(cos y − 1)3
Problem 4. Consider the functions
f (x) = 2x3 − 15x2 + 24x on [0, 5]
1. Find the critical points of f on the given interval.
2. Determine if they are local maximum, local minimum or neither.
3. Determine the absolute extreme values of f on the given interval when they exist.
solution. f 0 (x) = 6x2 − 30x + 24 = 6(x2 − 5x + 4) = 6(x − 4)(x − 1). So the critical
points are x = 1, 4.
By 1st derivative test, x = 4 are local minimum and x = 1 are local maximum.
Notice that f (4) = 128 − 240 + 96 = −16, f (1) = 11, f (0) = 0 and f (5) =
250 − 375 + 120 = −5. So absolute minimum are 4 and absolute maximum are
1.
Problem 5. A normal line at a point P on a curves define by the equation
√
x 3 y + 2y = 18
passes through P = (1, 8) and is perpendicular to the line tangent to the curve at P .
Determine an equation of the normal line at the given point.
1. Verify that the given P lies on the curve.
2. Determine an equation of the normal line of the curve at the given point.
solution. First, by implicit differentiation, we obtain that
√
3
y+
x −2/3 0
y
y + 2y 0 = 0 .
3
That is,
1 0
y (1) + 2y 0 (1) = 0,
12
hence y 0 (1) = −24/25. Then the slope for the normal line is −1/y 0 (1) = 25/24 and an
equation for the normal line is
2+
y−8=
25
(x − 1).
24
Problem 6. Consider the following functions on the given interval. Find the inverse
function, express it as a function of x, and find the derivative of the inverse function.
2
f (x) =
√
x + 2, for x ≥ −2
solution.
y=
√
x+2 .
implies that
x = y2 − 2 .
So
f −1 (x) = x2 − 2
and
(f −1 (x))0 = 2x .
Problem 7. The following limits represent the derivative of a function f at a point a.
Find a possible f and a, and then evaluate the limit.
sin2 π4 + h − 12
lim
h→0
h
solution. Let f (x) = sin2 x
π
4
sin2
+h −
lim
h→0
h
f π4 + h − f
= lim
h→0
h
0 π
= f
4
1
2
π
4
Now
f 0 (x) = 2 sin x · cos x
and
f
0
π 4
√
√
2
2
=2·
·
=1
2
2
Problem 8. A rectangle is constructed with its base on the x-axis and two of its vertices
on the parabola y = 12−x2 . What are the dimensions of the rectangle with the maximum
area? What is the area?.
solution. Let us assume the coordinate of the four vertices of the rectangle is given by
(−x, y), (x, y), (−x, 0), (x, 0) with the vertices (±x, y) is on the curve y = 12 − x2 .Then
the height of the rectangle is y and (±x, y) and x2 + y = 12.
We want to maximise the area A = 2xy = 2x(12 − x2 ). To do that, we solve
A0 (x) = 24 − 6x2 = 0
which implies that x = 2 that is, y = 8 and area is 2xy = 2 · 2 · 8 = 32.
Problem 9 ( points). Let f (x) = x3 − 3x
1. Is f (x) even, odd or neither? Explain why.
solution. It is odd function since f (−x) = −x3 + 3x = −f (x).
2. Find f 0 (x) and f 00 (x).
3
solution. f 0 (x) = 3x2 − 3 = 3(x + 1)(x − 1) and f 00 (x) = 6x.
3. Find critical points and possible inflection points.
solution. Critical point x = ±1 and inflection point x = 0 since f 00 (x) changes sign
at 0.
4. Find the intervals on which the function is increasing/decreasing and concave
up/down.
solution. f (x) concave down on (−∞, 0] and concave up on [0, +∞).
5. Identifying local maximum, local minimum and inflection points.
solution. By 2nd derivative test, we have x = −1 is local maximum since f 00 (−1) =
−6 < 0 and x = 1 is local maximum since f 00 (1) = 6 > 0.
6. Find x-intercept and y-intercept of the graph of f (x).
solution.
√ f (x) = x(x −
and ± 3.
√
3)(x +
√
3) then the roots for f (x) = 0 is given by x = 0
4