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201-NYA-05 May 2010 Final Exam
(1) Refer to the following sketch (with unit lengths marked along the coordinate axes) to answer the questions below.
x
y
b c
b c
b c
b
(a) Evaluate the following. Use ∞, −∞ or “does not exist” as appropriate. (i) lim x→− 2 −^
f (x) (ii) lim x→− 1
f (x) (iii) lim x→−∞
f (x) (iv) f (2) (v) lim x→ 1
f (x) (vi) lim x→ 2 f (x) (vii) lim x→ 4 f (x) (viii) lim x→∞ f (x) (b) List the values of x at which f is discontinuous. (c) List the values of x at which f is continuous but not dif- ferentiable. (2) Evaluate the following limits. Use ∞, −∞ or “does not exist” as appropriate.
(a) lim x→− 2
x^3 − 4 x 3 x^2 + 7x + 2 (b) lim x→∞
2 x^3 − 2 x^2 + 4x − 1 8 − 5 x + 3x^2 − 3 x^3 (c) lim ϑ→ 0
2 ϑ + 3 −
sin ϑ
(d) lim x→ 3
x 3 x − 2 x − 3 (3) (a) State the definition of the derivative of a function f. (b) Use the definition to find the derivative of f (x) = x/(x − 2). (c) Check your answer to Part b using the laws of differenti- ation. (4) Find all values of c such that
f (x) =
3 c x^2 − 10
if x < −4, and √ c − 2 x if − 4 ≤ x ≤ −2. is continuous at −4. (5) Find an equation of each line which is tangent to the graph of y = x^2 + 3x + 4 and passes through the point (2, 5).
(6) Sketch the graph of f (x) = x^2 /^3 (x − 3), given that f ′(x) = 5 x − 6 3 x^1 /^3
and f ′′(x) = 10 x + 6 9 x^4 /^3
. Make sure your solution includes all intercepts, asymptotes, intervals of monotonicity and concavity, extrema, and points of inflection. (7) Find all absolute extrema of: (a) f (x) = x^1 /^3 e^3 x^ on [ − 1 , 1 ]; (b) g(x) = 2 cos x sin x − 2
on [ − 12 π, 12 π ]. (8) Find all critical numbers of the function y = (4x + 3)^3 (3x + 3)^4. (9) Find the dimensions of the rectangle of largest area which has two vertices on the x-axis and two vertices above the x-axis, bounded by the curve y = 16 − 2 x^2. (10) For each of the following, find dy dx
(a) y = x^7 (x^3 − 2 x sin 3x)^2 /^3 (b) y = 5xπ^ − 5 /x +
x^2 − log 5 x + 5 · 2 x (c) y = ln
tan 2x^4 (d) y = sec
cos(tan πx)
(e) xy^2 + y ln x = x (11) Use logarithmic differentiation to find dy dx
(a) y =
sec^2 (5x − 2) x^12 e−x (b) y = (cos x)^5 x (^2) + (12) Find an equation of the normal line to the curve x^3 y − 2 y^3 − x + 3y = −11 at the point (− 1 , 2). (13) Let f (x) = e^2 x. Find a simplified formula for f (n)(x). (14) A ladder 12 metres long is leaning against a wall. The top does not reach high enough so the bottom of the ladder gets pushed towards the wall at a rate of 10 cm/s. What is the rate of change of the acute angle between the ladder and the floor when the ladder reaches 8 metres up the wall? (15) Find f (x) given that f ′′(x) = 4 − 6 x − 4 x^3 , f (1) = 2 and f ′(−1) = 1. (16) For the integral
1
x
dx, approximate its value with a Rie- mann sum with n = 6 rectangles using midpoints as sample points. (17) Evaluate the following integrals.
(a)
1
(3x + 1)^2 dx
(b)
(2x^ + sin x + π) dx
(c)
9 π 2 1 9 π 2
cos
x dx
(d)
x^3 + x^2 + x + 1 x
dx
(e)
3 π 0
sin x cos^2 x
dx
(18) Find the derivative of the function f (x) =
x^2
1 + t^2. (19) Sketch the graph of
f (x) =
2 x − 2 if x ≤ 3, and
4 if x > 3.
and evaluate
0
f (x) dx by interpreting it in terms of area.
Answers and/or Solutions:
(1) (a) By inspecting the given sketch: (i) lim x→− 2 −^
f (x) = ∞; (ii) lim x→− 1 f (x) = 2; (iii) lim x→−∞ f (x) = 0; (iv) f (2) = 1; (v) lim x→ 1 f (x) = ∞; (vi) lim x→ 2 f (x) does not exist, because lim x→ 2 −^
f (x) = 3 and lim x→ 2 +^
f (x) = −2; (vii) lim x→ 4
f (x) = −5; (viii) (^) xlim→∞ f (x) = −1. (b) f has infinite discontinuities at −2 and 1, a removable discontinuity at −1, and a jump discontinuity at 2. (c) f is continuous but not differentiable at 4, since lim t→ 4 −
f (t) − f (4) t − 4
= − 32 and lim t→ 4 +
f (t) − f (4) t − 4
(2) (a) Factoring the numerator and denominator and simplify- ing, gives
lim x→− 2
x^3 − 4 x 3 x^2 + 7x + 2
= lim x→− 2
x(x − 2)(x + 2) (3x + 1)(x + 2)
= lim x→− 2
x(x − 2) 3 x + 1
(b) Extracting the dominant powers of x from the numerator and denominator gives
lim x→∞
2 x^3 − 2 x^2 + 4x − 1 8 − 5 x + 3x^2 − 3 x^3
= lim x→∞
2 − 2 /x + 4/x^2 − 1 /x^3 8 /x^3 − 5 /x^2 + 3/x − 3
(c) Rationalizing the numerator, and using the basic limit lim ϑ→ 0
sin ϑ ϑ = 1, gives
lim ϑ→ 0
2 ϑ + 3 −
sin ϑ
lim ϑ→ 0
sin ϑ ϑ
· lim ϑ→ 0
2 ϑ + 3 +
(d) Simplifying the complex rational expression in the limit gives
lim x→ 3
x 3 x − 2 x − 3
= lim x→ 3
2(x − 3) 7(3x − 2)(x − 3)
= lim x→ 3
7(3x − 2)
(3) (a) The derivative f ′^ of a function f is defined by
f ′(x) = lim h→ 0
f (x + h) − f (x) h or, equivalently,
f ′(x) = lim t→x
f (t) − f (x) t − x
and the domain of f ′^ is the set of all real numbers x such that this limit exists. (b)
f ′(x) = lim t→x
t t − 2
x x − 2 t − x
Simplifying this expression, and applying the indepen- dence and direct substitution properties of limits then gives
f ′(x) = = lim t→x
t(x − 2) − x(t − 2) (t − 2)(x − 2)(t − x)
= lim t→x
−2(t − x) (t − 2)(x − 2)(t − x)
= lim t→x
(t − 2)(x − 2)
= −
(x − 2)^2
and the domain of f ′^ is equal to the domain of f (namely R \ { 2 }). (4) Observe that (since we must have c ≥ −4 for f to be defined on [ − 4 , −2 ]) lim x→− 4 −^
f (x) = 12 c and f (−4) = lim x→− 4 +^
f (x) =
c + 8.
So f is continuous at −4 if, and only if, c = 2
c + 8. This equa- tion implies that c^2 = 4(c+8), or 0 = c^2 − 4 c−32 = (c−8)(c+4), i.e., c = 8 or c = −4. However, only 8 is a solution of the original equation (c = −4 turns the original equation into the contradiction −4 = 4). Therefore, f is continuous at −4 if, and only if, c = 8. (5) The equation of the line tangent to the graph of y = x^2 + 3x + 4 at the point where x = ξ has slope 2ξ + 3, equation y = ξ^2 + 3ξ + 4 + (2ξ + 3)(x − ξ), and passes through the point (2, 5) if, and only if 5 = ξ^2 + 3ξ + 4 + (2ξ + 3)(2 − ξ), or 0 = ξ^2 − 4 ξ − 5 = (ξ − 5)(ξ + 1), i.e., ξ = 5 or ξ = −1, where the slope of tangent is, respectively 2(5)+3 = 13 and 2(−1)+3 = 1. Therefore, the line y = 5+13(x−2) = 13x−21, which is tangent to the parabola at (5, 44), and the line y = 5 + (x − 2) = x + 3, which is tangent to the parabola at (− 1 , 2), each pass through (2, 5), as does no other line tangent to the parabola. (6) The domain of f (x) = x^2 /^3 (x − 3) is R, on which f is continuous, so its graph has no vertical asymptotes. Since f (x) = x^5 /^3 (1 − 3 /x) for x 6 = 0, f (x) → ±∞ as x → ±∞, and the graph of f has no horizontal or oblique asymptotes, or global extrema. f (x) = 0 if x = 0 or x = 3, so the origin and (3, 0) are the intercepts of the graph of f. Since f (x) = x^2 /^3 (x − 3) = x^5 /^3 − 3 x^2 /^3 , one has f ′(x) = 53 x^2 /^3 − 2 x−^1 /^3 = 13 x−^1 /^3 (5x − 6) for x 6 = 0, so the critical numbers of f are 0 and 65 , f ′(x) > 0 if x < 0 or x > 65 and f ′(x) < 0 if 0 < x < 65. Therefore, f is increasing on ( −∞, 0 ) and on ( 65 , ∞ ), and decreasing on ( 0, 65 ), with a local maximum at the origin and a local minimum at ( 65 , − 2593
Next, f ′′(x) = 109 x−^1 /^3 + 23 x−^4 /^3 = 29 x−^4 /^3 (5x + 3) for x 6 = 0, so f ′′(x) < 0 if x < − 35 and f ′′(x) > 0 if − 35 < x < 0 or x > 0. Therefore, f is concave down on ( −∞, − 53 ), and con- cave up on ( − 53 , 0 ) and on ( 0, ∞ ), with a point of inflection at ( − 35 , − 18253
45). Since
lim t→ 0 ±
f (t) − f (0) t − 0
= lim t→ 0 ±
t^2 /^3 (t − 3) t
= lim t→ 0 ±^
t−^1 /^3 (t − 3) = ∓∞,
it follows that the graph of f has a vertical cusp at the origin. Below is a sketch of the graph of f , with the points of interest emphasized.
x
y = x^2 /^3 (x − 3)
b b
b b
5 ,^ −^
18 25
(7) (a) If f (x) = x^1 /^3 e^3 x, then f ′(x) = 13 x−^2 /^3 e^3 x^ + x^1 /^3 e^3 x^ 3 = 13 x−^2 /^3 e^3 x(1 + 9x) for x 6 = 0. Therefore, the critical numbers of f in [ − 1 , 1 ] are − (^19) and 0. Evaluating f at these critical numbers and at
(e) Revising the integrand and using a standard integral for- mula gives ∫ (^1) 3 π 0
sin x cos^2 x
dx =
3 π 0
sec x tan x dx = sec x
1 3 π 0
(18) First, note that f (x) = −
∫ (^) x 2
0
1 + t^2 dt. Then, by the Chain Rule and the (First) Fundamental The- orem of Calculus, f ′(x) = − 2 x
1 + x^4
(19) Here is a sketch of the graph of f , with the region representing the integral shaded.
x
y = f (x)
The region below the x-axis is a triangle with base 1, height 2, and area is 1. The region above the x-axis decomposes natu- rally into a triangle with base 2, height 4, and area 4, and a rectangle with base 3, height 4, and area 12. Therefore, ∫ (^6)
0
f (x) dx =
0
f (x) dx+
1
f (x) dx+
3
f (x) dx = −1+4+12 = 15.
