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M151B Practice Problems for Final Exam
Calculators will not be allowed on the exam. Unjustified answers will not receive credit. On the exam you will be given the following identities:
∑^ n
k=
k =
n(n + 1) 2
∑^ n
k=
k^2 =
n(n + 1)(2n + 1) 6
∑^ n
k=
k^3 =
(n(n + 1) 2
- Compute each of the following limits:
1a. lim x→ 2 −
x x^2 + 3x − 10
1b. lim x→ 0
xesin(^
(^1) x ) .
1c.
xlim→ 0
x sin x (1 − ex)^2
1d. lim x→∞
[(x + 1)^1 /^3 − x^1 /^3 ].
1e. The geometric mean of two positive real numbers a and b is defined as
ab. Show that
√ ab = lim x→∞(
a^1 /x^ + b^1 /x 2
)x.
2a. Find a value for c that makes the given function continuous at all points.
f (x) =
sin x x ,^ x^6 = 0 c, x = 0
2b. Determine whether or not your function from (2a) is differentiable at x = 0. If it is differentiable at this point, compute its derivative there.
- Find an equation for the line that is tangent to the graph of
f (x) =
xex 1 + x^2
at the point x = 0.
- Suppose the angle of elevation of the Sun is decreasing at a rate of .25 rad/hr. How fast is the shadow cast by a 400 ft tall building increasing when the angle of elevation of the Sun is π 6?
- Suppose f (x) is continuous on the interval [a, b] and differentiable on the interval (a, b). Show that if f ′(x) = x for all x ∈ (a, b), then there exists some value c ∈ (0, 1) so that
f (b) − f (a) = c(b − a).
- Let f (x) = x^1 /^3 (x + 3)^2 /^3 , −∞ < x < ∞.
6a. Locate the critical points of f and determine the intervals on which f is increasing and the intervals on which f is decreasing.
6b. Locate the possible inflection points for f and determine the intervals on which f is concave up and the intervals on which it is concave down.
6c. Evaluate f at the critical points and at the possible inflection points, and determine the boundary behavior of f by computing limits as x → ±∞.
6d. Use your information from Parts a-c to sketch a graph of this function.
- Find the side-lengths that maximize the area of an isosceles triangle with given perimeter P = 10. (An isosceles triangle is a triangle with two sidelengths equal.)
- Find all fixed points for the recursion equation
an+1 =
an +
an
Sketch a graph of the function f (a) = 34 a+ (^1) a , and use the method of cobwebbing to determine whether or not one of these fixed points will be achieved from the starting value a 0 = 12.
- Find all fixed points for the recursion equation
xt+1 = 1 +
xt
and determine whether or not each is asymptotically stable or unstable.
- Suppose a function f (x) is continuous on the interval [0, 1] and that you are given the following table of values:
x f (x) 1 / 8 1 / 2 3 / 8 1 / 3 5 / 8 − 1 7 / 8 − 2
Table 1: Values of f (x) for Problem 1.
Use an appropriate Riemann sum to approximate
0 f^ (x)dx.
- Use the method of Riemann sums to evaluate ∫ (^2)
1
x + x^2 dx.
1c. We apply L’Hospital’s Rule twice,
lim x→ 0
x sin x (1 − ex)^2
= lim x→ 0
sin x + x cos x 2(1 − ex)(−ex)
= lim x→ 0
sin x + x cos x 2 e^2 x^ − 2 ex
= lim x→ 0
2 cos x − x sin x 4 e^2 x^ − 2 ex^
1d. This limit has the indeterminate form ∞ − ∞, so the first thing we do is rearrange it into an expression with the form 00. We have
xlim→∞(x^ + 1)^1 /^3 −^ x^1 /^3 = lim x→∞ x^1 /^3 [(1 +
x
)^1 /^3 − 1]
= lim x→∞
(1 + (^) x^1 )^1 /^3 − 1 x−^1 /^3
= lim x→∞
1 3 (1 +^
1 x )
x^2 ) −^13 x−^4 /^3
= lim x→∞(1 +
x
)−^2 /^3
x^2 /^3
1e. We observe that this limit has the general form 1∞, and so we can apply L’Hospital’s rule. We have
xlim→∞(
a^1 /x^ + b^1 /x 2
)x^ = lim x→∞ eln(^
a^1 /x+ 2 b^1 /x)x = lim x→∞ ex^ ln(^
a^1 /x+ 2 b^1 /x)
= elimx→∞^ x^ ln(^
a^1 /x+ 2 b^1 /x) .
In order to compute this limit, we write
xlim→∞ x^ ln(
a^1 /x^ + b^1 /x 2
) = lim x→∞
ln(a
1 /x+b 1 /x 2 ) 1 x
= lim x→∞
2 a^1 /x^ +b^1 /x^ (
1 2 a
1 /x(ln a)(− 1 x^2 ) +^
1 2 b
1 /x(ln b)(− 1 x^2 )) − (^) x^12
= lim x→∞
a^1 /x^ + b^1 /x^
(a^1 /x^ ln a + b^1 /x^ ln b) =
(ln a + ln b),
where in this last step we have used that (^1) x → 0 as x → ∞. The limit is
e
(^12) (ln a+ln b) = e
(^12) ln(ab) = eln(ab)
1 / 2
ab.
2a. Since
xlim→ 0
sin x x
we can make this function continuous at all points by choosing c = 1.
2b. Since the function is separately defined at x = 0, we must proceed from the definition of differentiation. We compute
f ′(0) = lim h→ 0
f (0 + h) − f (0) h
= lim h→ 0
sin h h −^1 h = lim h→ 0
sin h − h h^2
= lim h→ 0
cos h − 1 2 h
= lim h→ 0
− sin h 2
where the last two steps both used L’Hospital’s rule. We conclude that this function is differentiable at x = 0, and that f ′(0) = 0.
- First,
f ′(x) =
(1 + x^2 )(ex^ + xex) − xex(2x) (1 + x^2 )^2
⇒ f ′(0) = 1,
which is the slope of the tangent line. Using f (0) = 0 and the general point-slope form y − f (a) = f ′(a)(x − a), we conclude y = x.
- First, observe that what we know is dθ dt = −.25 rad/hr and what we want to know is dx dt , where x is the length of the shadow (see the diagram).
We see that the relation between θ and x is
tan θ =
x
Upon taking a derivative of this equation with respect to t, we obtain
sec^2 θ
dθ dt
x^2
dx dt
where we can now fix θ = π 6 , so that sec^2 θ = (^) cos^12 π 6 = (^13) 4
= 43 , while x = (^) tan^400 π 6 = 400
Combining these observations, we have
dx dt
x^2 400
dθ dt
sec^2
π 6
= +400 ft/hr.
- According to the Mean Value Theorem there exists some value c ∈ (a, b) so that
f (b) − f (a) b − a
= f ′(c).
In this case f ′(c) = c, and so we conclude
f (b) − f (a) b − a
= c ⇒ f (b) − f (a) = c(b − a).
By the Pythagorean Theorem, the height of such a triangle is h =
y^2 − 14 x^2 , and so the area to be maximized is
A =
x
y^2 −
x^2 ⇒ A(x) =
x
x)^2 −
x^2 =
x
25 − 5 x, 0 ≤ x ≤ 5.
(The upper limit of 5 is clear both because a value of x larger than this would put a neg- ative number under the radical, and because the single side cannot be more than half the perimeter.) In order to maximize A(x), we compute
A′(x) =
25 2 −^
15 √^4 x 25 − 5 x
The critical values are x = 103 , 5, where we observe that x = 5 is also a boundary value. Checking A(x) at the critical and boundary values, we find
A(0) = 0
A(
A(5) = 0.
We conclude that the maximum area is 325 √ 3 and the side-lengths are x = 103 and y =
5 − 12 (^103 ) = 103. That is, an equilateral triangle.
- The fixed points solve
a =
a +
a
a =
a
⇒ a^2 = 4.
We conclude that the fixed points are ±2. In order to use cobwebbing, we must sketch a graph of the function
f (a) =
a +
a
First, setting
f ′(a) =
a^2
we find that the critical points are a = ± √^23 , 0. The function is increasing on (−∞, − √^23 ] ∪
[ √^23 , ∞) and decreasing on [− √^23 , √^23 ]. Next,
f ′′(a) =
a^3
and so the only possible point of inflection is a = 0. The function is concave down on (−∞, 0)
and concave up on (0, ∞). Finally,
a→−∞lim (
a +
a
f (−
lim x→ 0 −
a +
a
lim x→ 0 +
a +
a
f (
a→−∞lim (
a +
a
The plot of this function and the cobwebbing are depicted below. We conclude
lim n→∞
an = 2.
- In order to find the fixed points, we solve
x = 1 +
x
which becomes (upon multiplication by x)
x^2 − x − 2 = (x − 2)(x + 1) = 0,
and the fixed points are x = − 1 , 2. In order to check for stability we set f (x) = 1 + (^2) x , and compute
f ′(x) = −
x^2
12c. In this case, integrate by parts with u = cos−^1 x and dv = dx, for which we have du = − √ 11 −x 2 dx and v = x. The integral becomes
x cos−^1 x +
√ x 1 − x^2
dx.
For the remaining integral, we use fast substitution (since u has already been used) to obtain
x cos−^1 x −
1 − x^2 + C.
13a. We make the substitution u = 1 + x^3 (or alternatively use fast substitution), so that du = 3x^2 dx, and the integral becomes ∫ (^28)
2
x^2 √ u
du 3 x^2
2
u−^1 /^2 du =
u^1 /^2 1 / 2
28 2
[
2].
13b. We integrate by parts, setting
u = x dv = sec^2 xdx du = dx v = tan x.
We obtain ∫ π 4
0
x sec^2 xdx = x tan x
π 4 0
∫ π 4
0
tan xdx
π 4
π 4 0
π 4
- First, we locate the points of intersection by solving
x^4 = 20 − x^2 ⇒ x^4 + x^2 − 20 = 0.
In general, fourth order equations are difficult to solve algebraically, but this is really a second order equation in the variable x^2 , and it factors as
(x^2 − 4)(x^2 + 5) = 0,
so that the real roots are x = ±2. We observe that the upper graph is always y = 20 − x^2 , and also take advantage of symmetry to compute the area as
A = 2
0
(20 − x^2 ) − x^4 dx = 2[20x −
x^3 3
x^5 5
]^20 = 2[40 −
] =
- We observe that the graph of y = ex^ is always above the graph of y = e−x^ on [0, 2], and so according to the method of washers,
V = π
0
(ex)^2 − (e−x)^2 dx = π
0
e^2 x^ − e−^2 xdx
π 2
[e^2 x^ + e−^2 x]
2 0
π 2
[e^4 + e−^4 − 2],
where in this case we used fast substitution.
- First, the base of the triangle extends from y = x^2 up to y = x, so its length is x − x^2. The angle opposite the base is a right angle, so if we drop a line perpendicular to the base we divide the triangle into two 45-45-90 triangles. In this way, we see that the height of the triangle is x−x
2 2.^ (Alternatively, observe that the halves of the triangle can be rearranged into a square with sidelength x− 2 x 2 .) The area is
A(x) =
bh =
(x − x^2 )
x − x^2 2
(x − x^2 )^2.
The volume is
V =
0
(x − x^2 )^2 dx =
0
x^2 − 2 x^3 + x^4 dx =
[
x^3 3
x^4 2
x^5 5
]^10 =
