Math 114 Practice Midterm PDF

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MATH 114, D1

Midterm Examination, Version 2, Solutions October 19, 2016 Time: 50 minutes Chief Exam Administrator: E. Osmanagic

Name: Student ID: Email:

@ualberta.ca

Please note that this exam will be marked electronically; any writing on the back of pages will be lost and cannot be marked. • Scrap paper is supplied. • This is a closed book exam. No notes or books are permitted. • All electronic equipment, including calculators, is prohibited. Make certain that cell phones are turned off. • Be precise, concise, and use correct terminology in your answers.

IF ANYTHING IS UNCLEAR, PLEASE ASK!

Long Answer Problems. Unsubstantiated work may not receive full credit. 1. Differentiate. Do not simplify your result. p 5 2 a) (4 points) y = cot5 (3x) + 10 + ex −3x √ 3 b) (4 points) y = (3x3 − 3x)3 x5 + 3x c) (4 points) y =

cos3 (x2 ) (x3 + 5x)2

[12] Solution: 5 −3x2

a) y′ = 5 cot4 (3x) · (− csc2 (3x)) · 3 + (1/2)(10 + ex b) y′ = 3(3x3 − 3x)2 (9x2 − 3) ·

c) y′ =

5 −3x2

)−1/2 · (ex

) · (5x4 − 6x)

√ 3 x5 + 3x + (3x3 − 3x)3 (1/3)(x5 + 3x )−2/3 (5x4 + 3x ln 3)

3 cos2 (x2 ) · (− sin(x2 )) · 2x · (x3 + 5x)2 − cos3 (x2 ) · 2(x3 + 5x)(3x2 + 5) (x3 + 5x)4

Long Answer Problems. Unsubstantiated work may not receive full credit. 2

2. (a) (5 points) If g(2) = 1, g ′ (x) = g(x)ex for all x, find g ′′ (2). 1 (b) (5 points) Find the domain of the function f (x) = + ln(2x − 3x2 ). Write your 2x − 1 answer in the interval notation. [10] Solution: (a) (5 points) 2

2

g ′′ (x) = ex 2xg(x) + ex g ′ (x) For x = 2

2

2

g ′′ (2) = e2 4g (2) + e2 g ′ (2) 2

Since, g ′ (2) = e2 g(2) = e4 , we get 2

2

g ′′ (2) = e2 4g(2) + e2 g ′ (2) = 4e4 + e8 . (b) (5 points) Df = {x ∈ R | 2x − 3x2 > 0 and 2x − 1 6= 0}. 2x − 3x2 > 0 ⇐⇒ x(2 − 3x) > 0 ⇐⇒ x ∈ (0, 2/3). 2x − 1 6= 0 ⇐⇒ x ∈ (−∞, 1/2) ∪ (1/2, ∞). Hence, Df = (0, 1/2) ∪ (1/2, 2/3).

Long Answer Problems. Unsubstantiated work may not receive full credit. 3. Let f (x) =

x . x+3

a) ( 8 points) Use only the definition of the derivative to find f ′ (x). No marks will be given if the definition is not used. [8] Solution: f (x + h) − f (x) = lim h h→0 h→0

f ′ (x) = lim

x+h x+h+3

−

h

x x+3

= lim

h→0

(x + h)(x + 3) − x(x + h + 3) = h(x + 3)(x + h+)

x2 + 3x + xh + 3h − x2 − xh − 3x 3 3h = = lim . h→0 h→0 h(x + 3)(x + h + 3) h(x + 3)(x + h + 3) (x + 3)2 lim

b) (4 points) Find the point on the curve y = 1 +4ex − 6x where the tangent line is parallel to the line 6x − y = 3. Solution: Since y = 6x − 3, the slope of the given line is m = 6. On the other hand, the slope of the tangent line to the curve at x is y′ = 4ex − 6. Thus, 4ex − 6 = 6 ⇐⇒ 4ex = 12 ⇐⇒ ex = 3 ⇐⇒ x = ln 3. This implies, y(ln 3) = 1 + 4eln 3 − 6 ln 3 = 13 − 6 ln 3. Hence, the required point on the curve is P (ln 3, 13 − 6 ln 3).

[4]...