Math 1210-1 Midterm 2
Solution
1. Suppose f (x) = x3 − 3x + 1 defined on the closed interval [−2, 3].
(a) (6 points) Find all critical points.
Solution: f 0 (x) = 3x2 − 3 = 3(x − 1)(x + 1).
Stationary points: Solving f 0 (x) = 0, we have two stationary points
x = −1 and x = 1.
Endpoints: There are two endpoints: x = −2 and x = 3.
Singular points: Since f 0 (x) is defined everywhere, there is no singular point.
To conclude, all critical points are −2, −1 , 1 and 3 .
(b) (4 points) Find the maximum value and the minimum value.
Solution: Plugging in the critical points and finding the largest one
and the smallest one, we have f (−2) = −1, f (−1) = 3, f (1) = −1
and f (3) = 19. So the maximum value is 19 obtained at x = 3 and
the minimum value is −1 obtained at x = −2 and x = 1.
2. Suppose f (x) = 2x3 − 6x + 1 defined on all real numbers.
(a) (5 points) Find the interval where f (x) is decreasing.
Solution: f 0 (x) = 6x2 − 6. Since that f (x) is decreasing is equivalent
to f 0 (x) < 0, solving the inequality f 0 (x) < 0, we have −1 < x < 1 .
(b) (5 points) Find the interval where f (x) is concave up.
Solution: f 0 (x) = 12x. Since that f (x) is concave up is equivalent to
f 00 (x) > 0, solving the inequality f 00 (x) > 0, we have x > 0 .
3. (10 points) Suppose f (x) = x4 − 4x3 . Find the local extremes.
Solution: Since f 0 (x) = 4x3 − 12x2 = 4x2 (x − 3), the critical points are
x = 0 and x = 3. Checking the sign of f 0 (x) in each interval, we find that
f 0 (x) < 0 when x < 0 or 0 < x < 3 and f 0 (x) > 0 when x > 3. So by the
first derivative test, x = 0 is neither a local maximum nor a local minimum
and x = 3 is a local minimum .
Remark: If you want to run the second derivative test, you will find
f 00 (x) = 12x2 − 24x and f 00 (0) = 0. This fact won’t tell you that x = 0 is
not a local extreme. It only tells you that you can not apply the second
derivative test and have to use the first derivative test.
1
4. (10 points) Evaluate the following definite integral:
Z √π
2
x · sin(x2 ) dx
0
du
Solution: Let u = x2 . Then du = 2xdx, hence dx =
. Noticing that
2x
r
π
π
when x = 0, u = 0 and that when x =
, u = , we have
2
2
√
Z π2
Z π2
du
2
x · sin(x ) dx =
x · sin u
2x
0
0
Z π2
1
=
sin u du
2 0
π
1
= [− cos u]02
2
1
= (0 − (−1))
2
=
1
.
2
5. (10 points) Find the following derivative of definite integral:
Z 0
d
1
dt
dx tan x 1 + t2
Simplify your answer to get 2 bonus points.
sin x
.
Hint: tan x =
cos x
Solution: By the variant form of First Fundamental Theorem of Calculus,
we have
Z 0
d
1
dx tan x 1 + t2
1
1
=
· (0)0 −
· (tan x)0
1 + 02
1 + (tan x)2
1
1
= 0−
·
2
1 + tan x cos2 x
1
= − 2
cos x + tan2 x cos2 x
1
= − 2
cos x + sin2 x
=
−1 .
2
Z
6. Suppose
Z
2
Z
3
f (x) dx = 1,
3
g(x) dx = −1 and
0
g(x) dx = 1.
0
Z
2
2
(a) (4 points) Evaluate
g(x) dx.
0
Solution: By interval additivity,
Z 2
Z 3
Z
g(x) dx =
g(x) dx −
0
0
Z
(b) (6 points) Evaluate
3
g(x) dx = −1 − 1 = −2
2
2
(5f (x) − 3g(x) + 1) dx.
0
Solution: By linear property,
Z 2
Z
(5f (x) − 3g(x) + 1) dx = 5
0
Z
2
0
Z
2
f (x) dx − 3
2
g(x) dx +
0
1 dx
0
2
= 5 · 1 − 3 · (−2) + [x]0 = 13
7. (10 points) The graph of g(x) is given in the following picture.
(a) g(x) is increasing on the intervals [ a ,
b ] and [ d ,
(b) g(x) is concave up on the interval [ c ,
e ].
(c) The maximal value is achieved at x = b .
The local maximal value is achieved at x = b
(d) The inflection points are x =
(e) The singular point is x =
c
and x =
e .
e .
You don’t have to show your work in this problem.
3
and x =
e ].
e .
x
and f (1) = 2.
f (x)
Solution: Separating the variable and taking the integral, we have
8. (10 points) Find the function f (x) such that f 0 (x) =
Z
f 0 (x)f (x) = x
Z
f 0 (x)f (x) dx = x dx
f 2 (x)
x2
+ C1 =
+ C2
2
2
f 2 (x) = x2 + 2(C2 − C1 )
Let C = 2(C2 −C1 ) since they are both constants. We have f 2 (x) = x2 +C.
Plugging in f (1) = 2 when x = 1, we get 22 = 12 + C or C = 3. So
√
f 2 (x) = x2 + 3 or f (x) = ± x2 + 3. Since f (1) = 2 > 0, we should take
√
the positive square root. Hence, f (x) = x2 + 3 .
9. (10 points) Simplify the following sum:
n
X
(i + 1)(i + 2)
i=2
Solution: Noticing that the sum is taken from the second term to the nth
term, we can not apply the special sum directly. Since the special sum is
taken from the first term to the nth term, the difference is just the first
term. So we add the first term to the sum then subtract it.
n
X
(i + 1)(i + 2) =
n
X
(i + 1)(i + 2) − (1 + 1)(1 + 2)
i=1
i=2
=
n
X
(i2 + 3i + 2) − 6
i=1
=
n
X
i2 + 3
i=1
=
n
X
i=1
i+
n
X
2−6
i=1
n(n + 1)(2n + 1)
n(n + 1)
+3·
+ 2n − 6
6
2
4
10. (Bonus 10 points) Suppose









f (x) =








2
DNE
2x
DNE
1
x2
when
when
when
when
x<0
x=0
0<x<1
x=1
when x > 1
Find the unique continuous function F (x) such that F 0 (x) = f (x) and
F (−1) = 1.
Hint 1: Since f (x) is given piece by piece, your answer should also be
written piece by piece.
Hint 2: Sketching a graph of F (x) will help a lot.
Solution: Since F 0 (x) = f (x), F (x) must be one of the antiderivatives of
f (x) in each part. That is,


2x + C1 when x < 0


x2 + C2 when 0 < x < 1
F (x) =


 − 1 + C3 when x > 1
x
We are going to find proper values of C1 , C2 , C3 , F (0) and F (1) such that
F (x) is continuous and F (−1) = 1.
First, since −1 < 0, we plug F (−1) = 1 into the expression F (x) = 2x+C1
when x < 0. So 1 = 2(−1) + C1 , hence C1 = 3.
By the requirement that F (x) is continuous, we must have lim F (x) =
x→0
F (0). So
F (0) = lim− F (x) = lim− 2x + 3 = 3
x→0
x→0
We have F (0) = 3.
On the right limit,
3 = F (0) = lim+ F (x) = lim+ x2 + C2 = C2
x→0
x→0
So C2 = 3.
By the continuity of F (x) again, at x = 1 we must have lim F (x) = F (1).
x→1
So
F (1) = lim F (x) = lim x2 + 3 = 4
x→1−
x→1−
We have F (1) = 4.
5
Finally, on the right limit,
4 = F (1) = lim+ F (x) = lim+ −
x→1
x→1
1
+ C3 = −1 + C3
x
So C3 = 5.
To conclude,


2x + 3





3


x2 + 3
F (x) =


4





 −1 + 5
x
6
when
when
when
when
x<0
x=0
0<x<1
x=1
when x > 1.