Calculus 1 Assignment 6 Solutions
Alex Cowan
[email protected]
1.
a) The critical points, our candidates for mins/maxs, are where (sin(x))0 = 0. Since (sin(x))0 = cos(x), the critical
5π
points are ± π2 , ± 3π
2 , ± 2 , etc. Moreover, cos(x) is positive at 0, ±2π, ±4π, ... and negative at ±π, ±3π, ±5π, ....
Since cos(x) is continuous, the intermediate value theorem guarantees that it won’t change signs between zeros.
This means that at π2 , our function sin(x) goes from increasing to decreasing, and so must have a local max
there. By the same reasoning, it has a local max at ±2π + π2 , ±4π + π2 , ±6π + π2 , etc. It has local mins at
3π
3π
±2π + 3π
2 , ±4π + 2 , ±6π + 2 , etc. At these points, sin(x) is always 1 (for the maxs) or −1 for the mins, so
all these local mins/maxs are also global mins/maxs.
b) The critical points will be the same as above, so ( π2 , 1) (which is the only critical point in this interval)
is a local and global max. There are no mins at all.
c) In addition to the critical points where cos(x) = 0, there will be critical points wherever sin(x) = 0 as
the derivative won’t exist there (from the left the slope will be −1 and from the right the slope will be 1).
Finally, −4 and 4 are critical points, since we have to be able to take two sided limits for the derivative to exist.
At − π2 and π2 (the zeroes of cos(x) in this interval), | sin(x)| = 1. These are local and global maxs, since we
know that sin(x) is between −1 and 1 always.
At −π and π (the zeroes of sin(x)), | sin(x)| = 0. These are local and global mins (since |anything| ≥ 0).
At −4, the function sin(x) (without absolute values) is positive and decreasing (since −2π < −4 < 2 − π),
and the absolute value won’t change it (since the absolute value of a positive number is just that same number),
so −4 is a local max.
At 4, the function sin(x) (without absolute values) is negative and decreasing. The absolute value flips it
across the x axis, so | sin(x)| is positive and increasing there. That makes 4 a local max. Neither 4 nor −4 is
going to be a global max, since we know where | sin(x)| = 1, and it’s not at ±4.
d) The critical points are at −2 (where the function is not differentiable) and wherever f 0 (x) = 12x3 + 12x2 = 0.
We can factor 12x3 + 12x2 = 12x2 (x + 1), so its zeroes are at −1 and 0.
At −2, we can check that f0 (−2) < 0, so the function is decreasing. That means that −2 will be a local max.
At −1: Check that f 0 − 12 > 0. This means that to the right of −1, f is increasing. Since it’s decreasing to
the left of −1, that means that −1 will be a local min.
At 0: Clearly 3x4 + 4x3 is increasing for positive x. Since f (x) is increasing just to the left of 0 as well, we
conclude that 0 is not a local min or a local max.
As x gets close to 5, the function f (x) is large and positive. This means that there will be no global max,
and that (−1, −1) is a global min.
e) The derivative of x1 is −1
x2 . This function is never 0, and is defined everywhere
ical points, and hence no local/global mins/maxs.
1
x
is, so there are no crit-
2. a) Assume that (b, f (b)) is not a global max. We’ll show that this implies the existence of a critical point
(and this is enough to prove the statement in the problem).
If (b, f (b)) is not a global max, then there must be some point c ∈ (a, b) such that f (c) > f (b). This means
1
that
f (b) − f (a)
f (b) − f (c)
> 0 , and
<0
b−a
b−c
(x)
Since it was assumed that f is continuous, the function f (b)−f
is also continuous. The inequalities above mean
b−x
that this function goes from positive to negative, so the Intermediate Value Theorem implies that there is some
(y)
= 0. The Mean Value Theorem then implies that there is a point z (in
point y (in (a, c)) such that f (b)−f
b−y
0
(y, b)) such that f (z) = 0, i.e. there is a critical point somewhere.
b) The function (1 − x) sin x1 is an example. It oscillates very quickly as it approaches x = 0, and each peak is
higher and higher. No peak can ever be a global max, because the next one over is higher, and similarly for the
mins.
3.
q
a) f (x) = 0 for x = 0 or whenever 5 − 3x2 = 0. This happens when x = ± 53 .
q
Between −∞ and − 53 , the function is positive (since f (−10000) > 0, and IVT guarantees the function
q
can’t change sign between −∞ and − 53 ).
q
Between − 53 and 0, the function is negative (since f (−1) < 0).
q
q
Between 0 and 53 the function is positive, and past 53 the function is negative (since f is odd, e.g.).
f 0 (x) = 15(x2 − x4 ) = 15x2 (1 − x)(1 + x). This is zero for x = −1, 0, 1.
Between −∞ and −1, the function is decreasing.
Between −1 and 0, the function is increasing.
Between 0 and 1, the function is increasing, and between 1 and ∞, the function is decreasing (since f 0 is an
even function, e.g.).
q
f 00 (x) = 30(x − 2x3 ) = 30x(1 − 2x2 ). This is zero for x = 0, ± 12 .
q
Between −∞ and − 12 , f 00 (x) > 0, so f is concave up there.
q
Between − 12 and 0, the function is concave down.
q
Between 0 and 12 , the function is concave up.
q
Between 12 and ∞, the function is concave down.
Putting this together, we can sketch the function:
b) The derivatives of f are f 0 (x) = 1 − cos(x) and f 00 (x) = sin(x). Doing the same thing as above, we find
2
that f (0) = 0 and that f is always positive, that f is increasing everywhere, f 0 (x) = 0 for x = 0, 2π, 4π, and is
concave up for (0, π), concave down for (π, 2π), concave up for (2π, 3π), concave down for (3π, 4π). This gives
us the graph
3
2
6
12x
16x
00
c) f 0 (x) = x44x
+27 and f (x) = x4 +27 − (x4 +27)2 .
f (x) is always positive.
f 0 (x) = 0 for x = 0 only.
f 00 (x) = 0 when 12x2 (x4 + 27) − 16x6 = 0. This happens when either x = 0 or −4x4 + 12 · 27 = 0, which
happens when x = ±3.
All together, f is always positive, decreasing for x < 0 and increasing for x > 0, and concave up on (−3, 3)
and concave down otherwise. That means it looks like this:
4. The way I think about this is at every point I ask myself “Is the function increasing or decreasing” (which
would make the derivative positive (above the x axis) or negative (below the x axis)), and “Is it changing quickly”
(which would make the derivative be larger in absolute value, i.e. farther from the x axis.).
3
5.
a) Differentiate x3 + y 3 − 6xy + 3 = 0 with respect to t to get
dx
dy
dy
dx
3x2
+ 3y 2
−6 x
+y
=0
dt
dt
dt
dt
At t = 1, we’re given that x = 1, y = 2, and
dx
dt
= 5. This lets us solve for
dy
dt
at t = 1. We get
2 dx
6y dx
dy
dt − 3x dt
= 7.5
=
dt
6x − 3y 2
Then we use linear approximation: x(1.01) ≈ x(1) + 0.01
dx dt t=1
= 1.05 and y(1.01) ≈ y(1) + 0.01
dy dt t=1
3
= 2.075.
If we plug our guesses for x(1.01) and y(1.01) into the equation for the curve, we get x(1.01) + y(1.01)3 −
6x(1.01)y(1.01) + 3 ≈ 0.02.
dx
b) If we had gotten dy
dt = 5 and dt = 7.5 instead, then we would guess x(1.01) ≈ 1.075 and y(1.01) ≈ 2.05.
Plugging this in gives about −0.37 (which is much bigger!).
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