MATH 1210-1: Quiz 5A
Solution
Please show your work on parts (b), (c), (d) and (e) of the problem and MARK
your answer.
1. Confider the function f (x) = x2 − 2x on the closed interval [−2, 2].
(a) (1 point) f 0 (x) = 2x − 2 .
(b) (3 points) Find all three types of critical points.
Solution:
i. Stationary Points: Solve the equation f 0 (x) = 0 which is 2x−2 =
0 and get the solution x = 1.
ii. Endpoints: Obviously, x = −2 and x = 2.
iii. Singular Points: Since f 0 (x) is defined everywhere on [−2, 2],
there is no singular point.
Combining the three types together, we get that there are three critical points and they are −2, 1 and 2 .
(c) (2 points) When x =
−2
, f (x) reaches its maximum
value on [−2, 2], and the maximum value is
8
.
Solution: All the candidates of maximum value are the critical points
and the maximum value is the largest value achieved by the critical
points. Since f (−2) = 8, f (1) = −1 and f (2) = 0, we get that when
x = −2 , f (x) achieves its maximum value 8 on the closed interval
[−2, 2].
(d) (1 points) Find the average slope of f (x) on [−2, 2].
Solution: By definition,
average slope =
f (2) − f (−2)
0−8
=
= −2 .
2 − (−2)
4
(e) (3 points) Find the real number c ∈ [−2, 2] satisfying the Mean Value
Theorem of Derivatives.
Solution: By the Mean Value Theorem of Derivatives, there is a c
such that
f 0 (c) = average slope = −2.
Since f 0 (c) = 2c − 2, by solving the equation 2c − 2 = −2, we get
c=0.
MATH 1210-1: Quiz 5B
Solution
Please show your work on parts (b), (c), (d) and (e) of the problem and MARK
your answer.
1. Confider the function f (x) = x2 + 2x on the closed interval [−2, 2].
(a) (1 point) f 0 (x) = 2x + 2 .
(b) (3 points) Find all three types of critical points.
Solution:
i. Stationary Points: Solve the equation f 0 (x) = 0 which is 2x+2 =
0 and get the solution x = −1.
ii. Endpoints: Obviously, x = −2 and x = 2.
iii. Singular Points: Since f 0 (x) is defined everywhere on [−2, 2],
there is no singular point.
Combining the three types together, we get that there are three critical points and they are −2, −1 and 2 .
(c) (2 points) When x =
2
, f (x) reaches its maximum value
on [−2, 2], and the maximum value is
8
.
Solution: All the candidates of maximum value are the critical points
and the maximum value is the largest value achieved by the critical
points. Since f (−2) = 0, f (−1) = −1 and f (2) = 8, we get that
when x = 2 , f (x) achieves its maximum value 8 on the closed
interval [−2, 2].
(d) (1 points) Find the average slope of f (x) on [−2, 2].
Solution: By definition,
average slope =
f (2) − f (−2)
8−0
=
= 2.
2 − (−2)
4
(e) (3 points) Find the real number c ∈ [−2, 2] satisfying the Mean Value
Theorem of Derivatives.
Solution: By the Mean Value Theorem of Derivatives, there is a c
such that
f 0 (c) = average slope = 2.
Since f 0 (c) = 2c + 2, by solving the equation 2c + 2 = 2, we get
c=0.