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Math 112 โ Winter 2007 โ Key
Departmental Final Exam
Part I: Fill in the blank or circle T/F
- (a) The limit lim xโโ 2 x^2 + 3x 1 โ x^2
(b) If f (x) = x^2 , and x 0 = 1, then Newtonโs method for solving f (x) = 0 gives us x 1 =
(c) ( T / F ) If f โฒโฒ(x) exists on [a, b], then f (x) is continuous on [a, b]
(d) The mean value theorem states that if f is differentiable on [a, b], then there is a c in (a, b) with
f โฒ(c) = f (b) โ f (a) b โ a
(e) The limit lim xโ3+
|x โ 3 | x โ 3
(f) The average value of a function f over an interval [a, b] is given by
b โ a
โซ (^) b
a
f (x) dx
(g) If
2
f (x) dx = 2,
0
f (x) dx = 6,
0
g(x) dx = 5, then
โซ (^2)
0
f (x) + 3g(x) dx = 19
(h) ( T /F ) If f โฒ(x) exists on [a, b], then f (x) is integrable on [a, b].
(i) The integral
dx 1 + x^2 = tanโ^1 x + C
Part II: Multiple Choice
Problems 2 through 8 are multiple choice. Each multiple choice problem is worth 4 points. In the grid below fill in the square corresponding to each correct answer.
2 A C D E F G H I J
3 A B C E F G H I J
4 A B C D E F H I J
5 A B C D F G H I J
6 A B C D F G H I J
7 A B C D E G H I J
8 A B C D F G H I J
- lim xโ 0 2 x sin 5x
A. 0 B.
C. 1 D.
E. 2 F. โโ G. โ H. Limit does not exist.
- Given the limit statement lim xโ 5 (โ 3 x + 17) = 2, pick the largest ฮด that works with the definition
of the limit if = 0.06. A. 0.001 B. 0.005 C. 0.01 D. 0. E. 0.03 F. 0.06 G. 0 H. No such ฮด.
- If f (x) = 6x^2 , g(โ1) = โ 2 , gโฒ(โ1) = 3, find d dx (f (g(x))) at x = โ1. A. 0 B. 1 C. โ 12 D. โ 24 E. โ 36 F. โ 48 G. โ 72 H. None of the above.
- Which of the following is the maximum value of f (x) = 2x^3 โ 3 x^2 โ 36 x + 4 over [โ 3 , 2]?
A. 4 B. โ 2 C. 3 D. 80 E. 48 F. 64 G. โ 16 H. None of the above.
- Given x^2 ln y + y ln(x^2 ) = 2e, find dy dx at the point (
e, e).
A. โ 1 B. e C. โ 2 e D.
e 2
e โ 2 E. โ 2
e F. โ
2 e G. 1 H.
e + 2 โ e โ 2
Part III: Written Solutions
For problems 9 - 18, write your answers in the space provided. Neatly show your work for full credit.
- (a) State the conditions for f (x) defined over [0, 2] to be continuous at x = 1.
i. f (x) is defined at x = 1 ii. lim xโ 1 f (x) exists and is equal to f (1).
(b) At which points does the function
f (x) =
x + 4 (x + 2)(x โ 3)
fail to be continuous? At which points, if any, are the discontinuities removable? not removable? Give reasons for your answers.
The function f is not defined for all x < โ4 and at x = โ 2 , 3 and thus discontinuous at these points. All the discontinuous points are not removable because at these points the limit of f does not exist.
- Differentiate the following:
(a) f (x) = xฯ^ + sec(tan x) Applying power rule and chain rule, and recalling that
(sec x)โฒ^ = sec x tan x, (tan x)โฒ^ = sec^2 x,
f โฒ(x) = ฯ xฯโ^1 + sec(tan x) tan(tan x) sec^2 x
(b) g(x) =
x^2 + 1 x^2 โ 1
Applying chain rule and quotient rule
gโฒ(x) =
x^2 + 1 x^2 โ 1
x^2 + 1 x^2 โ 1
x^2 + 1 x^2 โ 1
โ 4 x (x^2 โ 1)^2
=
x^2 + 1 x^2 โ 1
โ 2 x (x^2 โ 1)^2
- Find the equation for the tangent line to f (x) = ex^ cos(x) at the point (ฯ, โeฯ).
f โฒ(x) = ex^ cos (x) โ ex^ sin (x) โ f โฒ(ฯ) = โeฯ so the equation for the tangent line at the point (ฯ, โeฯ) is given by
y + eฯ^ = โeฯ(x โ ฯ) or y = โeฯx + (ฯ โ 1)eฯ
- The area of a square is increasing at 4 in^2 /s. How fast is the length of the diagonal increasing at the moment that the side of the square is 6 in?
Let x the side of the square. Then the rate of increase of the area A = x^2 is given by
dA dt
in^2 /s. Now dA dt = 2x dx dt
dx dt
x The length of the diagonal is `(x) =
2 x, so the rate of increase of the diagonal is
d` dt
dx dt
x Hence when the side of the square is 6 in, the diagonal is increasing at a rate of
d` dt
(in/s)
- Given that for all x > โ4, the function f (x) is defined, continuous and satisfies the bounds
2 1 + eโ^1 /x^2 โค f (x) โค 2 + x 4 โ
x + 4
Determine the value f (0). State any theorem you used to find your answer.
As lim xโ 0
1 + eโ^1 /x^2
and lim xโ 0 2 + x 4 โ
x + 4
= lim xโ 0 2 + x(4 +
x + 4) 16 โ (x + 4)
so by the squeeze play or comparison limit theorem,
xlimโ 0 f^ (x) = 2. Since f (x) is continuous , so f (0) = 2.
- Let A(x) =
โซ โx
x+
sin t^2 dt. Find
dA dx
. State any theorem you used to find your answer.
By the fundamental theorem of calculus and the chain rule,
dA dx
d dx
โซ โx
x+
sin t^2 dt = sin
x)^2
) (^) d dx
x) โ sin
(x + 1)^2
) (^) d dx (x + 1)
and so dA dx
sin x 2
x โ sin
(x + 1)^2
- If f (x) =
x x^2 + 1
find all intervals of monotonicity, all intervals of concavity, all inflection points, all relative extrema and all global extrema if possible.
f โฒ^ = 1 โ x^2 (x^2 + 1)^2 , f โฒโฒ^ = 2 x(x^2 โ 3) (x^2 + 1)^3 so inflection points are x = 0, ยฑ
3 and extrema are x = ยฑ1.
x f โฒโฒ(x) f โฒ(x) f (x) โ 1 12 0 relative min 1 โ^12 0 relative max < โ 1 โ decreasing (โ 1 , 1) + increasing
1 โ decreasing < โ
3 โ conave down (โ
3 , 0) + conave up (0,
- โ conave down
3 + conave up The only relative maximum is also the global maximum. The only relative minimum is also the global minimum.
