School of Mathematics and Statistics
Junior
The University of Sydney
Maths & Stats Home / Teaching program / Junior / MATH1901 / Quizzes / Quiz 7
The University of Sydney
Usyd Home
MyUni
MyUni
Library
Sitemap
spcr
   Questions




right first
    attempt
right
wrong

Quiz 7: Differentiable functions

Last unanswered question  Question  Next unanswered question
 

Question 1

 
 
Find the equation of the tangent line to y = √ ------ e5x + 3 at x = 0.
a) x - 4y + 8 = 0   b) x - 4y - 8 = 0
c) 5x - 4y + 8 = 0   d) 5x - 4y + 4√ - 3 = 0

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
Not correct. Choice (d) is false.
 

Question 2

 
 
Find any critical points of the function f(x) = sinx + x on the interval [- 2π,2π] , and classify them.
a) The only critical point is a point of inflection at (π,π).
b) The only critical point is a minimum point at (π,π).
c) There are points of inflection at (-π,-π) and (π,π).
d) There are minimum points at (-π,-π) and (π,π).
e) There are no critical points.

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 3

 
 
Find the absolute minimum and maximum values of the function
f(x) = 2x3 - 3x2 - 12x + 45
 on the closed interval [-3,3].
a) The absolute minimum is 0, and the absolute maximum is 52.   b) The absolute minimum is 0, and the absolute maximum is 25.
c) The absolute minimum is 0, and the absolute maximum is 36.   d) The absolute minimum is 25, and the absolute maximum is 52.
e) The absolute minimum is 36, and the absolute maximum is 52.

 

Your answer is correct.
df(x) 2 dx = 6x - 6x- 12 = 0 when x = 2,-1. f(-1) = 52, f(2) = 25 and checking end points f(-3) = 0, f(3) = 36. So minimum is f(-3) = 0 and maximum is f(-1) = 52.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 4

 
 
Find the minimum and maximum values of f(x) = x3 - 9x + 8 on the interval [-3,1], if they exist.
a) 0 and 8   b) 0 and √- 8+ 6 3
c) √3-, and 8+ 6√3   d) 8 - 6√3 and 8 + 6√3-
e) √- 8 - 3 and √- 8 + 3

 

Not correct. Choice (a) is false.
Your answer is correct.
df(x) = 3x2 - 9 = 0 dx when x = - - √3 or - √ 3. However - √3 > 1 and so x = - √3 is not in the interval [-3,1]. Comparing f(-3) = 0, f(1) = -8 and f(-√- 3) = 8 + 6√- 3 gives the correct answer.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 5

 
 
Consider the function f(x) = |2x - 4| . Which of the following are true statements?
a) f is nowhere differentiable.   b) f is differentiable except at x = 2.
c) f is a discontinuous function.   d) df(x)= 2 dx for all x⁄=2.
e) df(x-)= - 2 dx for all x < 2.

 

There is at least one mistake.
For example, choice (a) should be false.
There is at least one mistake.
For example, choice (b) should be true.
There is at least one mistake.
For example, choice (c) should be false.
There is at least one mistake.
For example, choice (d) should be false.
There is at least one mistake.
For example, choice (e) should be true.
Your answers are correct
  1. False.
  2. True.
  3. False.
  4. False.
  5. True.
 

Question 6

 
 
According to the Mean Value Theorem, what is the largest number of real roots that the equation
x7 + 8x+ 13 = 0
can have?
a) 4   b) 1
c) 0   d) 2
e) 6

 

Not correct. Choice (a) is false.
Your answer is correct.
Let f(x) = x7 + 8x + 13. Then f′(x) = 7x6 + 8 , so f′(x) > 0 for all x. Since limx→ ±∞f (x) = ±∞ the function f(x) must have at least one root. If f(x) has two roots, say a < b, then f(a) = f(b) = 0. Applying the Mean Value Theorem of f(x) on the interval [a,b] we see that there must exist a number c ∈ [a,b] such that
f′(c) = f(b)--f(a)-= 0--0-= 0. b- a b- a
However, we have already seen that f′(x) > 0 for all x, so this is not possible. Hence, f(x) has exactly one root.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.
 

Question 7

 
 
Which of the following statements are true :
a) If f′(c) = 0, then f(x) has a maximum or minimum value at x = c.
b) If df(x) dg(x) --dx- = -dx-- for all x in an open interval I, then f(x) = g(x) on I.
c) If f(x) is differentiable on the open interval (a,b), and c is a point of local maximum for f in (a,b), then ′ f(c) = 0 .

 

There is at least one mistake.
For example, choice (a) should be false.
There is at least one mistake.
For example, choice (b) should be false.
There is at least one mistake.
For example, choice (c) should be true.
Your answers are correct
  1. False.
  2. False.
  3. True.
 

Question 8

 
 
If f(x) = 2x3 + x2 - x - 1, since f(x) is differentiable for all x, by the mean value theorem, there exists a c in [0,2] such that
′ f(2)- f(0) f (c) = ----2-----.
Find all possible values of c.
a) c = 1   b) √-- c = - 1-±-61- 6
c) c = ±1   d) - 1 - √61 c = ---6-----
e) √-- - 1-+-61- c = 6

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Your answer is correct.
Note that √ -- --1----61 6 is not in the interval [0,2].
 

Question 9

 
 
Let f (x) = |x - 1| . Then it is easy to see that there is no c ∈ [0,3] satisfying
f′(c) = f(3)--f(0). 3
This does not contradict the mean value theorem because (tick each correct answer):
a) f(x) is not continuous on [0,3].
b) f(x) is not differentiable on (0,3).
c) f(x) is not continuous everywhere.
d) The mean value theorem does not apply to functions with absolute values.

 

There is at least one mistake.
For example, choice (a) should be false.
f(x) is continuous for all x.
There is at least one mistake.
For example, choice (b) should be true.
f(x) is not differentiable at x = 1 and the mean value theorem only applies to functions which are differentiable on some closed interval.
There is at least one mistake.
For example, choice (c) should be false.
f(x) is continuous for all x.
There is at least one mistake.
For example, choice (d) should be false.
As long as the function is differentiable on some closed interval, the Mean Value Theorem applies, even if absolute values appear in the formula for the function. In particular, the mean value theorem does apply to the function f (x) = |x - 1| on, say, the closed interval [2,3], where it is differentiable.
Your answers are correct
  1. False. f(x) is continuous for all x.
  2. True. f(x) is not differentiable at x = 1 and the mean value theorem only applies to functions which are differentiable on some closed interval.
  3. False. f(x) is continuous for all x.
  4. False. As long as the function is differentiable on some closed interval, the Mean Value Theorem applies, even if absolute values appear in the formula for the function. In particular, the mean value theorem does apply to the function f (x) = |x - 1| on, say, the closed interval [2,3], where it is differentiable.
 

Question 10

 
 
Let f : ℝ → ℝ be a function that is differentiable everywhere. If f(1) = -3 and f′(x) > 5 , for all x, what is the best lower estimate for f(6) given by the mean value theorem?
a) f(6) > 22   b) f(6) > 25
c) f(6) < 25   d) f(6) > 8
e) f(6) < 22

 

Your answer is correct.
By mean value theorem
f(6)--f(1) ′ 5 = f (c) > 5
so f(6) > 22.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
Not correct. Choice (e) is false.