MA121, Homework #3
Solutions
1. Find the maximum value of f (x) = 3x4 − 16x3 + 18x2 over the closed interval [0, 3].
• Since f is continuous on a closed interval, we need only check the endpoints, the points
at which f 0 does not exist and the points at which f 0 is equal to zero. In our case,
f 0 (x) = 12x3 − 48x2 + 36x = 12x(x2 − 4x + 3) = 12x(x − 1)(x − 3)
and the only points at which the maximum value may occur are
x = 0,
x = 1,
x = 3.
Once we now note that the corresponding values of f at those points are
f (0) = 0,
f (1) = 5,
f (3) = −27,
we may finally conclude that the maximum value is f (1) = 5.
2. Show that the polynomial f (x) = x5 − 3x + 1 has three roots in the interval (−2, 2). As
a hint, you might wish to compute the values of f at the points ±2, ±1 and 0.
• Being a polynomial, f is continuous on the closed interval [−2, −1] and we also have
f (−2) = −25 < 0,
f (−1) = 3 > 0.
Thus, f must have a root in (−2, −1) by Bolzano’s theorem. Using the facts that
f (0) = 1 > 0,
f (1) = −1 < 0,
f (2) = 27 > 0,
we similarly find that a second root exists in (0, 1) and a third root exists in (1, 2).
3. Show that the polynomial f (x) = x5 − 5x + 1 has exactly one root in (1, 2).
• Being a polynomial, f is continuous on the closed interval [1, 2] and we also have
f (1) = −3 < 0,
f (2) = 23 > 0.
Thus, f has a root in (1, 2) by Bolzano’s theorem. Suppose now that f has two roots
in (1, 2). Then f 0 must also have a root in (1, 2) by Rolle’s theorem. Noting that
f 0 (x) = 5x4 − 5 = 5(x4 − 1) = 5(x2 + 1)(x − 1)(x + 1)
has no root in (1, 2), we deduce that f does not have two roots in (1, 2).
4. Suppose f is a differentiable function with |f 0 (x)| ≤ 1 for all x ∈ R. Show that
|f (x) − f (y)| ≤ |x − y|
for all x, y ∈ R.
• When x = y, the desired inequality states that 0 ≤ 0, so it is certainly true.
• Suppose now that x 6= y. According to the mean value theorem, we must have
f (x) − f (y)
= f 0 (c)
x−y
=⇒
|f (x) − f (y)| = |x − y| · |f 0 (c)|
for some c between x and y. Using the fact that |f 0 (c)| ≤ 1, we conclude that
|f (x) − f (y)| = |x − y| · |f 0 (c)| ≤ |x − y|.
5. Letting f (x) = x5 − 5x + 1 for all x ∈ R, compute each of the following:
max f (x),
−2<x<0
max f (x),
0<x≤2
min f (x).
−1≤x≤2
• To find the minimum and maximum values of f over an arbitrary interval, one needs
to determine the sign of f 0 throughout the interval. In our case, we have
f 0 (x) = 5x4 − 5 = 5(x4 − 1) = 5(x2 + 1)(x − 1)(x + 1)
and the sign of f 0 can be determined using the table below.
x
x +1
x−1
x+1
f 0 (x)
f (x)
2
−1
+
−
−
+
%
1
+
−
+
−
&
+
+
+
+
%
• According to the table, the maximum value of f (x) when −2 < x < 0 is f (−1) = 5.
• To find the maximum value of f (x) when 0 < x ≤ 2, we need to compare
f (0) = 1,
f (2) = 23.
Since the latter is bigger and also attained, this gives max f (x) = f (2) = 23.
0<x≤2
• According to the table, the minimum value of f (x) when −1 ≤ x ≤ 2 is f (1) = −3.