Download Math 1000 Tutorial Quiz 7 Week 7 The solutions to Quiz 7 are below

Math 1000 Tutorial
Quiz 7
Week 7
The solutions to Quiz 7 are below. I might be showing more steps than necessary, this is to aid your
understanding.
1
1. Find the critical numbers of f (x) = x4 + 3x2 + 6.
Solution: The derivative is given by f 0 (x) = 4x3 + 6x.
First, find when f 0 (x) = 0. If 4x3 + 6x = x(4x2 + 6) = 0, then either x = 0 or 4x2 + 6 = 0. There is
no real solution for the second, thus f 0 (x) = 0 only if x = 0.
Second, the derivative is defined everywhere.
Therefore, the critical number is x = 0.
2
2. Find the local and absolute maxima and minima of f (x) = x − x3 on [−1, 1].
Solution: (i) Find the critical numbers:
q
f 0 (x) = 1 − 3x2 , setting equal to zero we have x = ± 13 , both solutions are on the interval.
f 0 (x) is defined everywhere on the interval, thus these two are the only critical numbers.
(ii) Evaluate the function at the critical numbers and the endpoints of the interval:
f (−1) = 0
f (1) = 0
q
q
q
f ( 13 ) = 13 − ( 13 )3 > 0
q
q
q
f (− 13 ) = − 13 − (− 13 )3 < 0
(iii) Comparing these values we get:
q q
q
1
There is a local and absolute maximum at ( 13 , 13 − 27
).
q
q
q
1
There is a local and absolute minimum at (− 13 , − 13 + 27
).
1
3. (a) Does f (x) = x(x2 − x − 2) satisfy the criteria for the Mean Value Theorem on [−1, 1]? Explain why
or why not.
Solution: Because f (x) is a polynomial, it is continuous on [−1, 1] and differentiable on (−1, 1).
Thus f (x) satisfies the criteria for the Mean Value Theorem on [−1, 1].
1/
2
(bonus)
(b) Find a c ∈ (−1, 1) satisfying f 0 (c) =
f (1)−f (−1)
1−(−1)
for the above function.
Solution: By part (a) we can apply the Mean Value Theorem, thus such a c exists.
Now, f 0 (x) = 3x2 − 2x − 2, implying f 0 (c) = 3c2 − 2c − 2.
Also,
f (1)−f (−1)
1−(−1)
=
−2−0
2
= −1.
Thus setting these equal, we have 3c2 − 2c − 2 = −1. Using the quadratic formula we get
c = − 13 or c = 1.
The second solutions is not in the interval, thus the c satisfying the conclusion of the Mean
Value Theorem is c = − 13 in this case.