Math 31A Discussion Notes
Week 6
November 3 and 5, 2015
This week we’ll review two of last week’s lecture topics in preparation for the quiz.
Linearization
One immediate use we have for derivatives is local linear approximation. On small
neighborhoods around a point, a differentiable function behaves linearly. That is, if we zoom
in enough on a point on a curve, the curve will eventually look like a straight line. We can use
this fact to approximate functions by their tangent lines. You’ve seen all of this in lecture,
so we’ll jump straight to the formula for local linear approximation. If f is differentiable at
x = a and x is “close” to a, then
f (x) ≈ L(x) = f (a) + f 0 (a)(x − a).
Example. Use local linear approximation to estimate the value of sin(47◦ ).
(Solution) We know that f (x) := sin(x) is differentiable everywhere, and we know the
value of sin(45◦ ). Since 47◦ is reasonably close to 45◦ , this problem is ripe for local linear
π
cos(x), so f 0 (45◦ ) = 180π√2 . Then
approximation. We know that f 0 (x) = 180
1
π
90 + π
√ =
√ ≈ 0.7318.
sin(47◦ ) ≈ sin(45◦ ) + f 0 (45◦ )(47 − 45) = √ + 2
2
180 2
90 2
For comparison, Google says that sin(47◦ ) = 0.7314, so our estimate is pretty good.
Considering the fact that most folks now have (extremely powerful) calculators in their
pockets, the above example is a very inefficient way to compute sin(47◦ ). Local linear
approximation is no longer especially useful for estimating particular values of functions,
but it can still be a very useful tool. In a lot of applications it is convenient to assume that
certain functions are linear.1 The theory of local linear approximations allows us to do this
on relatively small neighborhoods.
Extreme Values
Next, we would like to identify the extreme values of continuous functions. As you saw in
lecture, we do this by first identifying the points at which the derivative of f vanishes or
does not exist, and then comparing the value of f at these points with the value of f at the
ends of its domain (either endpoints or end behavior).
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For example, the velocity of a falling object does not behave exactly linearly, but this nonlinear function
is well-behaved enough that we can approximate it with a linear function.
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Definition. A number c in the domain of a function f is called a critical point of f if
either f 0 (c) = 0 or f 0 (c) does not exist.
Theorem 1. Suppose f is continuous on the closed domain [a, b], and let c ∈ [a, b] be such
that f (c) is the minimum or maximum value of f on [a, b]. Then c is either a critical point
of f or one of the endpoints a or b.
Example. Find the maximum value of f (x) = x1/2 (1 − x)3 on the closed interval [0, 1].
(Solution) We have
1
f 0 (x) = x−1/2 (1 − x)3 + 3x1/2 (1 − x)2 (−1),
2
by the product and chain rules. Notice that f 0 (x) is defined on (0, 1], but not defined at 0.
Now if c ∈ (0, 1) is a critical point, then f 0 (c) = 0, so
1 −1/2
x
(1 − x)3 = 3x1/2 (1 − x)2
2
⇒
1/6(1 − c) = c
⇒
1
c= .
7
Now we must compare the value of f at c to the value of f at the endpoints:
r 3
1
1 6
,
f (1) = 0.
f (0) = 0,
f
=
7
7 7
Whatever the value of f (c), it is positive, and is thus the maximum value of f on [0, 1].
In some cases we can actually determine whether a critical point will give a local maximum
or a local minimum in a somewhat easier way, using the first derivative test for critical
points. For this, let c be a critical point of f . Then if f 0 (x) is positive just to the left of c
and negative just to the right of c, f (c) is a local maximum. On the other hand, if f 0 (x) is
negative just to the left of c and positive just to the right of c, then f (c) is a local minimum.
If f 0 (x) does not change signs at x = c, then f has neither a local maximum nor a local
minimum at x = c.
. Find the critical points of f and the intervals
Example. Consider the function f (x) = 2x+1
x2 +1
on which f is increasing or decreasing. Use the first derivative test to determine whether
the critical point(s) give local minima or maxima.
(Solution) First we use the quotient rule to find that
f 0 (x) =
(x2 + 1)(2) − (2x + 1)(2x)
−2x2 − 2x + 2
−2(x2 + x − 1)
=
=
.
(x2 + 1)2
(x2 + 1)2
(x2 + 1)2
Since (x2 + 1)2 > 0 for all x, f 0 (x) exists for all x. So if c is a critical point of f , then
f 0 (c) = 0. But this implies that
√
√
−1 − 5
−1 + 5
2
c +c−1=0
⇒
c1 =
and c2 =
.
2
2
2
So f has two critical points, one positive, one negative. We can find the intervals on which
f is increasing or decreasing by checking the sign of the derivative in the intervals with
endpoints at the critical points. In particular, −2 < c1 and f 0 (−2) < 0, so f must have
negative derivative at all points less than c1 . That is, f is decreasing on (−∞, c1 ). Similarly,
2 > c2 and f 0 (2) < 0, so f must have negative derivative to the right of c2 . So f is decreasing
on the interval (c2 , ∞). Finally, c1 < 0 < c2 and f 0 (0) > 0, so f is increasing on (c1 , c2 ).
Since f 0 is negative just to the left of c1 and positive just to the right of c1 , f (c1 ) is a local
min. On the other hand, f 0 is positive just to the left of c2 and negative just to the right of
c2 , so f (c2 ) is a local max for f .
Example. A marine energy company plans to place a tidal turbine in the ocean, 5km south
of a straight shore that runs east-west. The turbine must be connected via a power line to a
transformer that is located on the shore, 12km east of the point on the shore that is nearest
to the turbine. The cost of installing this power line is $30, 000 for each kilometer of line
that is in the water and $20, 000 for each kilometer along the shore. Assuming the line will
follow a straight line to the shore (which is not necessarily perpendicular to the shore) and
then follow the shore to the transformer, what is the minimum cost of installing the power
line?
(Solution) Let x represent the distance between the point where the power line hits the
shore and the point on the shore which is closest to the turbine. Then the part of the power
line that is in the water is the hypotenuse of a right triangle
√ with legs of length x and 5km,
so the length of the line which is in the water is given by 25 + x2 . The length of the part
of the line which runs along the shore is given by 12 − x, so the total cost of the line is given
by
√
c = (30) 25 + x2 + (20)(12 − x),
measured in thousands of dollars. Notice that c is defined on the closed interval [0, 12]. We
see that
30(2x)
dc
= √
− 20,
dx
2 25 + x2
p
30x
so c has a critical point when √25+x
=
20.
This
is
easily
solved
to
give
x
=
25/1.25 =
2
√
2 5 ≈ 4.47km. Finally, we notice that
√
√
c(0) = 390,
c(2 5) = 50 5 + 240 ≈ 351.803,
and
c(12) = 390.
So the most cost-effective plan for our power line is to reach the shore approximately 4.47km
east of the point on the shore which is nearest the turbine, and the installation of the power
line will cost approximately $351,803.
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