SOLUTION MANUAL THEORETICAL MODEL
SIMULATION METHODOLOGY DIDACTIC
ACCLAMATION CONSTRUCT 2026
⫸ 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) =. Ans: 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.. Ans: 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 = ...... Ans: 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)). Ans: 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
SIMULATION METHODOLOGY DIDACTIC
ACCLAMATION CONSTRUCT 2026
⫸ 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) =. Ans: 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.. Ans: 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 = ...... Ans: 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)). Ans: 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
