SOLUTION MANUAL EXAM WITH QUESTIONS AND ANSWERS VERIFIED GRADED A+

SOLUTION MANUAL EXAM WITH QUESTIONS AND ANSWERS
VERIFIED GRADED A+


Find the midpoint Riemann Sum of cos(x^2) with n = 4, from [0, 2] - ANSWER-Mid
S4 = (1)(1/2)[cos(.25^2) + cos(.75^2) + cos(1.25^2) + cos(1.75^2)]
Mid S4 = (1)(1/2)[cos(.625) + cos(.5625) + cos(1.5625) cos(3.0625)]
Mid S4 = .824

If the function f is continuous for all real numbers and if f(x) = (x^2-7x +12)/(x -4)
when x ≠ 4 then f(4) = - ANSWER-Factor numerator so
f(x) = (x-3)(x-4)/(x-4) = x-3
f(4)=4-3
f(4) = 1

If f(x) = (x^2+5) if x < 2, & f(x) = (7x -5) if x ≥ 2 for all real numbers x, which of the
following must be true?

I. f(x) is continuous everywhere.
II. f(x) is differentiable everywhere.
III. f(x) has a local minimum at x = 2. - ANSWER-At f(2) both the upper and lower
piece of the discontinuity is 9 so the function is continuous everywhere.

At f'(2) the upper piece is 4 and lower piece is 7 so f(x) is not differentiable
everywhere.

Since the slopes of the function on the left and right are both positive the function
cannot have a local minimum or maximum at x= 2.

Only I is true.

For the function f(x) = (ax^3-6x), if x ≤ 1, & f(x) = (bx^2+4), x > 1 to be continuous
and differentiable, a = ..... - ANSWER-for the function to be continuous f(1) has to
equal f(1):
a(1^3) -6(1) = b(1^2) +4
a -6 = b +4
b=a-10

for the functions to be differentiable f'(1) has to equal f'(1):
3a(1^2) -6 = 2b(1)
3a -6 = 2b
plug b from the first equation in to find a:
3a -6 = 2(a -10)
a = -14

, Find k if f(x) = (k) at x = 4 and f(x) = ((x^2 -16)/(x-4)) - ANSWER-1. f(4) exists and is
equal to 8
2. lim from the left and right are both 8
3. lim f(x) as x approaches 4 is 8 which equals f(4)

k must equal 8

If f(x) is continuous and differentiable and f(x) = (ax^4 +5x) for x ≤ 2, & f(x)= (bx^2 -3)
for x > 2 , then b =... - ANSWER-Plug x = 2 into both pieces.
f(x) = (16a +10) for x ≤ 2, & (4b -6) for x > 2
They must be equal to be continuous
16a +10 = 4b -6
a=.25b-1

Take the derivative of both pieces of this function and plug in x = 2
f(x) = (32a +5) for x ≤ 2, & f(x) = (4b -3) for x > 2
They must be equal to be differentiable
32a +5 = 4b -3
plug in the first equation to find b
32(.25b-1)+5= 4b-3

b=6

If f is continuous for a ≤ x ≤ b, then at any point x = c, where a < c < b, which of the
following must be true?

a. f(c) = (f(b) - f(a))/(b-a)
b. f(a) = f(b)
c. f(c) = 0
d. lim f(x) as x approaches c = f(c) - ANSWER-In order for f(x) to be continuous at
point c, there are three conditions that need to be fulfilled:

1. f(c) exists
2. lim f(x) as x approaches c exists
3. lim f(x) as x approaches c = f(c)
Answers a ,b , and c are not necessarily true.

If the function f(x) = (3ax^2+2bx +1), x ≤ 1, & f(x) = (ax^4-4bx^2-3x), x > 1 , is
differentiable for all real values of x, then b = ... - ANSWER-For the function to be
continuous both pieces must be equal at 1.
Plug in x = 1 into both pieces and set them equal to each other.
f(x) = (3a +2b +1), x ≤ 1, & f(x) = (a -4b -3), x > 1
3a +2b +1 = a -4b -3
2a +6b = -4.

For the function to be differentiable the derivative of both pieces must be equal at 1.
Plug x = 1 into the derivative of both pieces
f'(x) = (6a +2b), x < 1, & f'(x) = (4a -8b -3), x > 1
6a +2b = 4a -8b -3