Math 111
Quiz #9
Solutions
1. If f is differentiable, f (3) = 4, and f 0 (3) = 5, find
lim
x→0
f (3 + 2x) − f (3 − x)
.
x
We look at the limit as x approaches 0 of the numerator and the denominator separately to
see if this limit is one of the indeterminate forms. Since f is differentiable, f is also continuous, so
the numerator is continuous and hence taking the limit as x approaches 0 in the numerator amounts
to plugging in x = 0, so
lim (f (3 + 2x) − f (3 − x)) = f (3 + 0) − f (3 − 0) = 4 − 4 = 0.
x→0
The limit as x approaches 0 in the denominator is
lim x = 0.
x→0
So this limit is in the indeterminate form 00 . We apply L’Hospital’s rule, being careful to use the
chain rule where appropriate:
f (3 + 2x) − f (3 − x)
2f 0 (3 + 2x) − (−1)f 0 (3 − x)
= lim
.
x→0
x→0
x
1
lim
Since f is differentiable, f 0 must be continuous and so
lim 2f 0 (3 + 2x) + f 0 (3 − x) = 2f 0 (3 + 0) + f 0 (3 − 0) = 2f 0 (3) + f 0 (3) = 2(5) + 5 = 15.
x→0
Hence the original limit is
f (3 + 2x) − f (3 − x)
= 15.
x→0
x
√
2. Use pictures to show why Newton’s method fails when applied to the equation 3 x = 0 with any
initial approximation x0 6= 0.
lim
x3
x1
x0
x2
The successive x-intercepts Newton’s Method gives us will always be getting farther and
farther from the root x = 0. Since Newton’s Method only works if the x-intercepts of the tangent
lines eventually start getting closer and closer to the root and we can see that this will not happen,
Newton’s Method fails for this function and an initial guess x0 6= 0.
3. A fence 8 ft tall runs parallel to a very tall building at a distance of 4 ft from the building. What
is the length of the shortest ladder that will reach from the ground over the fence to the wall of the
building? [Hint: You will want to use the Pythagorean Theorem and similar triangles here.]
I didn’t intend for the calculations for this problem to be so difficult, so I graded only on if
you were able to set up the problem in a correct way and find an equation in one variable for the
quantity you wanted to optimize. It got messy after that point, and you wouldn’t be expected to
do such a messy problem on a test.
building
z
y
8
fence
4
ladder
x
From the picture we see that we have a right triangle involving the hypotenuse labeled by z, which
represents the length of the ladder. So from the Pythagorean Theorem we have the equation
z 2 = y 2 + (4 + x)2 .
We also have similar right triangles, one with vertical side 8 feet, the height of the fence, and the
other with vertical side y feet, where y indicates how far up the building the ladder touches. Using
the fact that ratios of sides for similar triangles are equal we have
y
x+4
=
.
8
x
We’ll solve this equation for y and substitute it in the equation above involving the length of the
ladder, z.
y
x+4
8(x + 4)
=
⇒ y=
,
8
x
x
and so
2
8(x + 4)
2
z =
+ (x + 4)2 .
x
At this point, most of you solved for z (since that’s what we want to minimize) and then took the
derivative. And the derivative is a mess. We can avoid most of the mess by observing that if z is
minimized, then the quantity z 2 will also be minimized. So consider the function
2
8(x + 4)
L(x) =
+ (x + 4)2
x
(which is z 2 , the square of the length of the ladder), and we’ll minimize this function.
8(x + 4) −32
0
L (x) = 2
+ 2(x + 4).
x
x2
There is a critical point at x = 0 (L’(0) DNE). This is an endpoint of our domain, since we want
the length of the piece labeled by x to have a positive length, and x = 0 corresponds to placing
the ladder vertically against the fence. In this case, the ladder will never reach the building (our
building is tall, but not that tall). So we solve for L0 (x) = 0.
8(x + 4) −32
2
+ 2(x + 4) = 0
x
x2
32 8(x + 4)
·
= (x + 4)
x2
x
As long as x 6= −4, we can divide both sides of this equation by the factor x + 4. And since the
domain that we are interested does not contain negative numbers, we assume that x 6= −4. So we
continue and get
32 8
· =1
x2 x
32 · 8 = x3
√
3
x = 256.
√
We have one critical point in the domain at x = 3 256 ≈ 6.35. We perform the first derivative test
to check that it is a minimum: L0 (4) = −48 and L0 (8) = 12 so
L’
-
+
4
6.35
8
√
So √L(x) has an absolute minimum when x = 3 256, and the absolute minimum value of L(x) is
L( 3 256) ≈ 277.15. But q
L(x) was the square of the length of the ladder, so the length of the shortest
√
ladder that will work is L( 3 256) ≈ 16.65.