Name:
Student number:
Math 110-003
Quiz 2 (Feb 4)
Write full solutions to the questions below. It is to your own benefit to write as clearly and
legibly as possible.
1. Find the critical numbers of the following functions:
(a) f (x) = 3x4 + 4x3 − 6x2
Solution.
We have
f 0 (x) = 12x3 + 12x2 − 12x = 12x(x2 + x − 1),
(1)
so f 0 (x) = 0 when x = 0 or when
q(x) = x2 + x − 1 = 0.
(2)
By the quadratic formula, q(x) = 0 for
−1 ±
x=
2
√
5
.
(3)
Since f 0 (x) is well-defined everywhere, the only critical numbers are those found
above:
√
−1 ± 5
.
(4)
0 and
2
t2 − 4
t2 + 4
Solution.
We have
(b) g(t) =
g 0 (t) =
2t(t2 + 4 − (t2 − 4))
16t
2t(t2 + 4) − (t2 − 4)(2t)
=
=
.
(t2 + 4)2
(t2 + 4)2
(t2 + 4)2
(5)
Since t2 ≥ 0, we have
(t2 + 4)2 ≥ (0 + 4)2 = 16 > 0,
(6)
i.e. the denominator of g 0 (t) is never 0. Thus, g 0 (t) is always well-defined and
g 0 (t) = 0 when the numerator (which is 16t) is 0, i.e. when t = 0. Thus, the only
critical number is 0.
Name:
Student number:
Math 110-003
Quiz 2 (Feb 4)
2. Consider the function h(x) = 4x5 + x3 + 2x + 1 on the interval [0, 1].
(a) Solution.
The function h(x) satisfies the hypotheses of the mean value theorem because it is
continuous on [0, 1] and differentiable on (0, 1).
(b) Show that there is a number c in (0, 1) such that h0 (c) = 7.
Solution.
By the mean value theorem (which holds by (a)), there is a number c in (0, 1)
where
h(1) − h(0)
= 4 + 1 + 2 + 1 − 1 = 7.
(7)
h0 (c) =
1−0