MATH 1210-1: Quiz 7A
Solution
Please show your work and MARK your answer.
NO credit will be given if you write down the solutions of the equations in Problem 1 and
2 directly, though they are easy to find.
1. (3 points) Use bisection method to find an approximation solution to the equation
x3 + x − 2 = 0. Let the starting interval be [0, 4]. Proceed two cycles of this method.
Solution: Let f (x) = x3 + x − 2. Noticing that f (0) = −2 < 0 and f (4) = 66 > 0, by
Intermediate Value Theorem, we have that the solution is indeed in the interval [0, 4].
Cycle 1. We pick the midpoint 2 of [0, 4]. Since f (2) = 8 > 0 who has different sign
from f (0), we conclude that the solution is in the interval [0, 2].
Cycle 2. We pick the midpoint 1 of [0, 2]. Since f (1) = 0, we conclude that x = 1
is the solution.
2. (4 points) Use Newton’s method to find an approximation solution to the equation
x3 + x + 2 = 0 and let x1 = 0. Find x2 and x3 . You don’t have to evaluate the fraction
in your answer.
Solution: By Newton’s Method, in order to solve the equation x3 + x + 2 = 0, we
f (xn )
are going to use the formula xn+1 = xn − 0
where f (x) = x3 + x + 2, hence,
f (xn )
f 0 (x) = 3x2 + 1. We have
x2 = x1 −
03 + 0 + 2
f (x1 )
=0−
= −2
0
f (x1 )
3 × 02 + 1
and
x3 = x2 −
(−2)3 + (−2) + 2
−8
18
f (x2 )
=
(−2)
−
= −2 −
= −
= −1.3846 . . .
f 0 (x2 )
3 × (−2)2 + 1
13
13
3. (3 points) Find the expression of the function F (x) such that F 0 (x) = x2 and F (1) = 2.
Z
Solution: By definition, F (x) is one of the antiderivatives of x2 , i.e., x2 dx. By the
x3
+ C. In order to find the expression of F (x),
3
x3
we only need to determine the value of C. Since F (1) = 2 and F (x) =
+ C, we
3
1
x3
5
5
have that 2 = + C or C = . Therefore, we conclude that F (x) =
+ .
3
3
3
3
power rule, this indefinite integral is
MATH 1210-1: Quiz 7B
Solution
Please show your work and MARK your answer.
NO credit will be given if you write down the solutions of the equations in Problem 1 and
2 directly, though they are easy to find.
1. (3 points) Use bisection method to find an approximation solution to the equation
x3 + x + 2 = 0. Let the starting interval be [−4, 0]. Proceed two cycles of this method.
Solution: Let f (x) = x3 + x + 2. Noticing that f (−4) = −66 < 0 and f (0) = 2 > 0,
by Intermediate Value Theorem, we have that the solution is indeed in the interval
[−4, 0].
Cycle 1. We pick the midpoint −2 of [−4, 0]. Since f (−2) = −8 < 0 who has the
same sign as f (−4), we conclude that the solution is in the interval [−2, 0].
Cycle 2. We pick the midpoint −1 of [−2, 0]. Since f (−1) = 0, we conclude that
x = −1 is the solution.
2. (4 points) Use Newton’s method to find an approximation solution to the equation
x3 + x − 2 = 0 and let x1 = 0. Find x2 and x3 . You don’t have to evaluate the fraction
in your answer.
Solution: By Newton’s Method, in order to solve the equation x3 + x − 2 = 0, we
f (xn )
are going to use the formula xn+1 = xn − 0
where f (x) = x3 + x − 2, hence,
f (xn )
f 0 (x) = 3x2 + 1. We have
x2 = x1 −
and
x3 = x2 −
f (x1 )
03 + 0 − 2
=
0
−
= 2
f 0 (x1 )
3 × 02 + 1
f (x2 )
23 + 2 − 2
8
18
=
2
−
=2−
=
= 1.3846 . . .
0
2
f (x2 )
3×2 +1
13
13
3. (3 points) Find the expression of the function F (x) such that F 0 (x) = x3 and F (1) = 2.
Z
Solution: By definition, F (x) is one of the antiderivatives of x3 , i.e., x3 dx. By the
x4
+ C. In order to find the expression of F (x),
4
x4
we only need to determine the value of C. Since F (1) = 2 and F (x) =
+ C, we
4
x4
1
7
7
have that 2 = + C or C = . Therefore, we conclude that F (x) =
+ .
4
4
4
4
power rule, this indefinite integral is