Download Analysis Homework #4 Solutions 1. 2.

Analysis Homework #4
Solutions
1. Find the values of x for which 6x3 + x < 5x2 .
• To find the values of x for which f (x) = 6x3 − 5x2 + x is negative, we write
f (x) = x(6x2 − 5x + 1) = x(2x − 1)(3x − 1)
and then worry about the signs of each of the factors separately.
x
x
2x − 1
3x − 1
f (x)
0
−
−
−
−
1/3
+
−
−
+
1/2
+
−
+
−
+
+
+
+
According to the table, f (x) is negative when x < 0 and also when 1/3 < x < 1/2.
2. Compute each of the following limits:
L = lim
x→+∞
1 + 2x − x2
,
3 − 2x + x2
M = lim
x→−∞
x3 − 2x + 1
.
x4 + 2x − 3
• Dividing both the numerator and the denominator by x2 , we easily get
L = lim
x→+∞
1 + 2x − x2
1/x2 + 2/x − 1
0+0−1
=
lim
=
= −1.
2
2
x→+∞ 3/x − 2/x + 1
3 − 2x + x
0−0+1
Dividing both the numerator and the denominator by x4 , we similarly get
M = lim
x→−∞
x3 − 2x + 1
1/x − 2/x3 + 1/x4
0−0+0
=
lim
=
= 0.
4
3
4
x + 2x − 3 x→−∞ 1 + 2/x − 3/x
1+0−0
3. Use the definition of the derivative to compute f ′ (y) when
f (x) = ax3 + bx2 + cx + d
for some constants a, b, c, d ∈ R.
• In this case, we have
f (x) − f (y)
a(x3 − y 3 ) + b(x2 − y 2 ) + c(x − y)
=
x−y
x−y
and we can simplify the expression on the right hand side to get
f (x) − f (y)
= a(x2 + xy + y 2 ) + b(x + y) + c.
x−y
Using the definition of the derivative, we now find
[
]
f (x) − f (y)
= lim a(x2 + xy + y 2 ) + b(x + y) + c
x→y
x→y
x−y
f ′ (y) = lim
and polynomials are known to be continuous, so
f ′ (y) = a(y 2 + y 2 + y 2 ) + b(y + y) + c = 3ay 2 + 2by + c.
4. Find the min and max values of f (x) = 3x4 − 16x3 + 18x2 over [−1, 1].
• Since f is continuous on a closed interval, it suffices to check the endpoints, the points
at which f ′ does not exist and the points at which f ′ is equal to zero. In this case,
f ′ (x) = 12x3 − 48x2 + 36x = 12x(x2 − 4x + 3) = 12x(x − 1)(x − 3)
so the only points at which the min/max values may occur are the points
x = −1,
x = 1,
x = 0,
x = 3.
We exclude the point x = 3, which does not lie in the given interval, and we note that
f (−1) = 37,
f (1) = 5,
f (0) = 0.
Thus, the minimum value is f (0) = 0 and the maximum value is f (−1) = 37.