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M151B Practice Problems for Exam 1
Calculators will not be allowed on the exam. Unjustified answers will not receive credit.
- Compute each of the following limits:
1a.
lim x→ 2
x 2 − 4
x − 2
1b.
lim x→ 3 −
x
x^2 − 2 x − 3
1c.
lim x→ 0
sin 7x
x
1d.
lim x→ 1
x^2 + 1 −
x + 1
x − 1
1e.
lim x→−∞
x^3 − x^2 + 1
1 − x^2
1f.
lim x→∞
(e −x sin x).
- Find all points at which
ln(1 − x)
ln(1 + x)
is continuous.
- Find a value for c that makes the given function continuous at all points.
f (x) =
x^2 + 1, x ≤ 1
x − c, x > 1
- Prove that the equation
e x − 2 = sin x
has at least one real-valued solution.
- Use the bisection method to approximate a root of
x 4
with a maximum error of 13.
- Use the definition of limits to prove the following statement:
lim x→ 2
7 x − 1 = 13.
- Use the definition of limits to prove the following statement:
lim x→ 3
(x 2
- Use the definition of derivative to compute the derivative of the following function at
x = 0.
f (x) =
x^2 cos( 1 x ),^ x^6 = 0 0 , x = 0.
- Determine whether or not each of the following functions is differentiable at the point
x = 0. In each case, explain why or why not.
9a.
f (x) =
x 2
x^2 − 1 , x > 0
9b.
f (x) =
x^2 + 1, x ≤ 0
2 x + 1, x > 0
9c.
f (x) = x|x|.
- Find an equation for the line that is tangent to the given curve at x = 1.
y = x 3
Sketch a graph of the curve along with this tangent line.
- Compute the derivative of each of the following functions:
11a.
f (x) = x
2 (^3) + x−^7.
11b.
f (x) = x sin x.
11c.
f (x) =
ex^ − e−x
ex^ + e−x^
11d.
f (x) = (2x +
x
2 .
Notice in particular that we don’t have to be able to evaluate the function at a point to
compute its limit at that point.
1b. Compute
lim x→ 3 −
x
x^2 − 2 x − 3
= lim x→ 3 −
x
(x − 3)(x + 1)
1c. We make the substitution y = 7x, and our limit becomes
lim y→ 0
sin y
(y/7)
= 7 lim y→ 0
sin y
y
You won’t lose points on a problem like this if you omit the explicit substitution.
1d. In this case, we rationalize the numerator,
lim x→ 1
x^2 + 1 −
x + 1
x − 1
= lim x→ 1
x^2 + 1 −
x + 1
x − 1
x^2 + 1 +
x + 1 √ x^2 + 1 +
x + 1
= lim x→ 1
x^2 + 1 − (x + 1)
(x − 1)(
x^2 + 1 +
x + 1)
= lim x→ 1
x^2 − x
(x − 1)(
x^2 + 1 +
x + 1)
= lim x→ 1
x(x − 1)
(x − 1)(
x^2 + 1 +
x + 1)
= lim x→ 1
x
(
x^2 + 1 +
x + 1)
1e. According to our rule from class, the following calculation is entirely fair:
lim x→−∞
x 3 − x 2
1 − x^2
= lim x→−∞
x 3
−x^2
= lim x→−∞
(−x) = +∞.
1f. Since sin x does not have a limit as x → ∞ we use the Squeeze Theorem (a.k.a. the
Sandwich Theorem), observing
−e −x ≤ e −x sin x ≤ e −x .
We have limx→∞(−e−x) = limx→∞(e−x) = 0, so by the Squeeze Theorem
lim x→∞
e −x sin x = 0.
- First, observe that ln(1 − x) is only defined for x < 1 and ln(1 + x) is only defined for
x > −1, so our range is restricted to this interval. Also, we cannot divide by 0, so we must
have x 6 = 0. We conclude that the points of continuity are
- We observe that the only point at which f may not be continuous is x = 1, and at this
point f (1) = 2. In order to make the function continuous at this point, we must ensure
lim x→ 1 +^
x − c = 2,
and this requires c = −1.
- We begin by defining the function
f (x) = e x − 2 − sin x,
and we note that our goal will be to show that f (x) has at least one real root. First, we
observe that f (0) = −1. Next, we observe that since e > 2 we know that e^2 > 4, so that
f (2) = e^2 − 2 − sin 2 > 0. We can conclude from the Intermediate Value Theorem that there
is a root on the interval (0, 2).
- We begin by defining the function
f (x) = x 4
and we observe that f (0) = −1 and f (1) = 2, so that we are guaranteed a root in (0, 1). We
take
c 1 =
and compute
f (
4
3
We conclude that the root is on the interval ( 1 2 ,^ 1), and our second approximation becomes
c 2 =
1 2 + 1 2
and 1 4 <^
1 3 , so this is a sufficient approximation.^ (Note.^ This equation has a second real root between -2 and -1, so it’s possible to approximate that one instead, which is fine.)
- Our goal will be to show that
|(7x − 1) − 13 | < ǫ.
For linear functions we choose δ = ǫ
- We now verify that this works by computing
|(7x − 1) − 13 | = | 7 x − 14 | = 7|x − 2 | < 7 δ = 7
ǫ
7
= ǫ.
This completes the proof.
- In this case we want to show
0 < |x − 3 | < δ =⇒ |(x^2 + 1) − 10 | < ǫ.
As usual, we begin with |(x 2
- − 10 |, and in this case we compute
|(x 2
- − 10 | = |x 2 − 9 | = |(x + 3)(x − 3)| = |x + 3||x − 3 |.
As our first choice of δ, we take δ = 1, so that
2 < x < 4 ,
while
lim h→ 0 +
(2h + 1) − 1
h
= lim h→ 0 +^
Since these limits do not agree, we can conclude that f (x) is not differentiable at x = 0.
Method 2. In cases for which
f (x) =
f 1 (x) x ≤ a
f 2 (x) x > a
where f 1 (a) = f 2 (a), f ′ 1 (a) =^ c^1 , and^ f^
′ 2 (a) =^ c^2 we can proceed as follows: if^ c^1 6 =^ c^2 then^ f
is not differentiable at x = a, while if c 1 = c 2 then f is differentiable at x = a and f ′(a) = c 1.
Here, f 1 (x) = x^2 + 1 and f 2 (x) = 2x + 1, so f ′(0) = 0 while f 2 ′(0) = 2. We can draw the
same conclusion as we did with Method 1. (Be sure to check all assumptions when using
this method; try it, for example, on (2a).)
9c. In this case,
f ′ (0) = lim h→ 0
h|h|
h
= lim h→ 0
|h| = 0,
and so f is differentiable at x = 0 with f ′(0) = 0.
- The slope of the tangent line is given by the derivative y′(1) = 3(1)^2 = 3. We have, then
y − 2 = 3(x − 1).
11a. Applying the power rule to each summand, we find
d
dx
(x
2 (^3) + x−^7 ) =
x − (^13) − 7 x − 8 .
11b. Applying the product rule, we find
d
dx
x sin x = sin x + x cos x.
11c. Applying the quotient rule, we find
d
dx
ex^ − e−x
ex^ + e−x^
(ex^ + e−x)(ex^ + e−x) − (ex^ − e−x)(ex^ − e−x)
(ex^ + e−x)^2
and though considerable simplification is possible, this form is sufficient for the exam.
11d. Proceeding with the chain rule, we set u = 2x + 1 x and compute
d
dx
u 2 =
d
du
u 2 du dx
= 2u(2 −
x^2
) = 2(2x +
x
x^2
(This substitution does not need to be made explicitly.)
- Method 1. Compute directly
f ′ (x) = cos(
2 x)
2 x^
x ln 2 =
ln 2
2
cos(
2 x)
2 x,
and
f ′′ (x) =
ln 2
2
− sin(
2 x)
ln 2
2
x
2 x)
ln 2
2
2 x
ln 2
2
)^2
2 x^ cos(
2 x) − 2 x^ sin(
2 x).
Method 2. First, observe that
2 x^ = (
2)x, which eliminates the need for a nested chain
rule. Now,
d
dx
sin((
x ) = cos((
x )(
x ln
and
d^2
dx^2
sin((
x ) = − sin((
x )((
x ln
2
x )((
x ln
- ln
= (ln
2
cos((
x )(
x − sin((
x ) x
which is equivalent to the expression from Method 1.
- We compute implicitly
d
dx
sin(xy) =
d
dx
x ⇒ cos(xy)
d
dx
(xy) = 1 ⇒ cos(xy)(y + x
dy
dx
Solving for
dy dx , we find dy
dx
1 cos(xy) −^ y x
At the point ( √^1 2
√ 2 π 4 ), we have
dy
dx
1 cos( π 4 ) −^
√ 2 π 4 √^1 2
π
2
The equation for the tangent line is
(y −
2 π
4
π
2
)(x −
The linear relationship is
log 10 y = (4 log 10 7)x + log 10 2.
- This is a semilog plot with log 10 y on the vertical axis and x on the horizontal. That is,
the line has an equation of the form
log 10 y = mx + b.
We can read directly from the plot that b = − 1. Likewise, the slope is
m =
We have, then,
log 10 y =
x − 1.
In order to get a functional relationship, exponentiate each side with base 10,
log 10 y = 10
1 2 x−^1 ⇒ y = 10−^1 (
1 (^2) )x.
