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(Marks) (5) 1. Use the graph of the function f (x) to find the following. Use ∞, −∞, or DNE where appropriate.
(a) f (−2) =
(b) f (2) = (c) (^) xlim→− 2 f (x) =
(d) lim x→ 0 f (x) =
(e) lim x→ 2 +^
f (x) =
(f) lim x→ 2 −^ f (x) =
(g) lim x→ 2 f (x) =
(h) (^) xlim→∞ f (x) =
(i) lim x→−∞ f (x) =
(j) The intervals on which f (x) is continuous.
x
y
− (^2 )
− 1
1
2
(15) 2. Evaluate the following limits. Use the algebraic methods whenever possible. Identify the limits that do not exist and use ∞ or −∞ where appropriate. Show your work.
(a) lim x→ 2
2 x+2 −^
1 2 x − 2
(b) lim x→ 2 3 x^2 − 5 x − 2 x − 2
(c) lim x→ 4 +
x^2 − 9 4 − x
(d) (^) x→−∞lim 8 x^3 + 3x − 7 5 x^2 + 12
(e) lim x→ 4
x + 32 − 6 x − 4 (4) 3. Find the point(s) of discontinuity of the function. Justify using the definition of continuity.
f (x) =
(x + 5)(x + 1)
if x < − 2
x + 1 5 if x ≥ − 2
(Marks) (3) 4. Find all the possible value(s) of k so that the function is continuous for all real numbers
f (x) =
3 x − 4 k if x ≥ 5 2 x + 9 if x < 5
(6) 5. (a) Use the limit definition of the derivative of a function to find f ′(x) for f (x) =
2 x + 5
(b) Check your answer to part a) using the appropriate rule of differentiation. (18) 6. Find y′^ for the given functions. Do not simplify your answer.
(a) y =
3 x − 1 2 x + 5
(b) y = 5x^2 + (^) x^52 +
x^5 − log 5 (2x + 1) − e^5
(c) y = 1 + csc x sin x (d) y = sec (cos(3x)) (e) y = etan^ x^ + ex^ cot(x) (f) y = (x^2 + 1)^2 x
(4) 7. Find d^2 y dx^2 for the function y = x^3 ex^2
(4) 8. Given the implicitly defines function xy = x^2 y − 3 x + y , find the slope of the tangent line at the point (0, 0).
(4) 9. Use the rules of logarithms to find the derivative of y = ln
(x^5 + 3)^2010 e^3 x^
6 − x^5 + 3x^2
(4) 10. Find an equation of the tangent line to the curve y = (x^2 + 1)
2 x − 3 at the point (2, 5).
(4) 11. Use the second derivative test to find the relative extrema of f (x) = x^3 − 9 x^2 + 24x − 20
(4) 12. Find the absolute extrema of f (x) = x^4 − 4 x^3 + 1 on the interval [− 1 , 4].
(10) 13. Given f (x) = x x^2 − 1 ; f ′(x) = −(x^2 + 1) (x^2 − 1)^2 ; f ′′(x) = 2 x(x^2 + 1) (x^2 − 1)^3
(a) List the x−intercept, y−intercept, equations of vertical and horizontal asymptotes (if any), relative extrema, points of inflection, intervals where f (x) is increasing, decreasing, concave up and concave down. (b) Sketch the graph of f (x).
(Marks) Answers
- a) 2 b) undefined c) − 1 d) − 5 e) +∞ f) − 1 g) D.N.E. h) 1
- i) −∞ j) x < −2 or − 2 < x < 2 or x > 2
- a) −^18 b) 7 c) −∞ d) −∞ e) 121
- x = −5 or x = − 2 4. k = − 1
- a) f ′(x) = lim h→ 0
f (x + h) − f (x) h b)f ′(x) =
(2x + 5)^2
- a) y′^ = 4
( 3 x− 1 2 x+
) 3 [3(2x+5)−2(3x−1) (2x+5)^2
]
b) y′^ = 2x
5 x^2 ln 5
− 10 x−^3 + 52 x^3 /^2 − (^) (2x+1) ln 5^2
c) y′^ = −^ csc^ x^ cot^ x.^ sinsin^ x 2 −x cos^ x(1+csc^ x) d) y′^ = −3 sin(3x) sec (cos(3x)). tan (cos(3x))
e) y′^ = sec^2 x. etan^ x^ + ex^. cot(x) − csc^2 (x). ex^ f) y′^ = (x^2 + 1)^2 x
[
2 ln(x^2 + 1) + 2 x x^2 + 1
. 2 x
]
d^2 y dx^2 = 2xex^2 (2x^4 + 7x^2 + 3) 8. y′(0, 0) = 3 9. y′^ = 2010 5 x 4 x^5 +3 −^3 −^
1 2
− 5 x^4 +6x 6 −x^5 +3x^2
- y = 9x − 13 11. relative maximum at (2, 0) and relative minimum at (4, −4)
- absolute maximum is 6 at x = −1 and absolute minimum is −26 at x = 3
- x-int and y-int:(0,0) vertical asymptotes: x = −1 and x = 1 horizontal asymptote: y = 0 no relative extrema; IP:(0, 0) Inc: never Dec: (−∞, −1) ∪ (− 1 , 1) ∪ (1, +∞) CU: (− 1 , 0) ∪ (1, +∞) CD: (−∞, −1) ∪ (0, 1)
- 15 additional trees
- a) R′(80000) = 30 If the production increases to 80001 units, the revenue will increase by about $30. b) P = − 0. 0005 x^2 + 70x − 5000 c) x = 70000 units
- a) η = 2(
√x−18) √x =⇒ η(100) = − 1. 6 b) elastic at x = 100 units c) unit elasticity at x = 144 units
f (x)
x
− (^1 ) 0
