School of Mathematics and Statistics
Junior
The University of Sydney
Maths & Stats Home / Teaching program / Junior / MATH1011 / Quizzes / Quiz 5
The University of Sydney
Usyd Home
MyUni
MyUni
Library
Sitemap
spcr
   Questions




right first
    attempt
right
wrong

Quiz 5: More on differentiation; maxima and minima; inflexions

Last unanswered question  Question  Next unanswered question
 

Question 1

 
 
Let f(x) = x3 - 3x2 + 7 . Which of the following statements are correct?
a) The graph of y = f(x) has only one turning point, at x = 2.   b) The absolute maximum value of f(x) is 7.
c) The graph of y = f(x) has a local maximum at x = 0.   d) The graph of y = f(x) has a local minimum at x = 0.
e) The graph of y = f(x) has one point of inflexion.

 

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 a local maximum value of 7, but it is not the absolute maximum.
There is at least one mistake.
For example, choice (c) should be true.
There is at least one mistake.
For example, choice (d) should be false.
There is a local maximum at x = 0.
There is at least one mistake.
For example, choice (e) should be true.
Your answers are correct
  1. False.
  2. False. There is a local maximum value of 7, but it is not the absolute maximum.
  3. True.
  4. False. There is a local maximum at x = 0.
  5. True.
 

Question 2

 
 
You are given that the function f(x) = 2(2x-+-3)- x + x- 2 has an absolute maximum on the interval -2 < x < 1. Find this maximum.
a) - 2 9   b) -10
c) -2   d) -5

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
f′(x) = - 2(x-+-5)(x-+-1) (x+ 2)(x- 1) . So the only critical value of f on the given interval occurs at x = -1. Given that f has a maximum it must be at x = -1. Moreover, we find that f(-1) = -2.
Note that f(x) is not defined at x = -2 or at x = 1, but f(x) approaches -∞ as x approaches 1 from below or -2 from above.
Not correct. Choice (d) is false.
 

Question 3

 
 
Find the absolute minimum of the function f(t) = t2 +-2t et on the interval -2 ≤ t ≤ 2.
a) 2- 2√2- ---√2-- e   b) √2-
c) √- 2-+√2-2 e 2   d) -√2

 

Your answer is correct.
2- t2 f′(x) = --et--
Critical points are at t = ±√2-. There is a minimum at t = -√2-.
√- 2- 2√2 f(- 2) = ---√2--≈ - 3.41 e
f(2) = 1.08, f(-2) = 0
therefore the absolute minimum is 2- 2√2 ---√2-- e at t = -√ - 2.
Not correct. Choice (b) is false.
This is a critical point.
Your answer is correct.
This is the absolute maximum.
Not correct. Choice (d) is false.
This is a critical point.
 

Question 4

 
 
Find the points of inflexion of the function 4 3 f(x) = x - 12x + 6x - 9 on the interval -2 ≤ x ≤ 10.
a) x = 0, 6   b) x = 0, -6
c) x = ±√6-   d) x = ±√12-

 

Your answer is correct.
f(x) = x4 - 12x3 + 6x - 9 f′(x) = 4x3 - 36x2 + 6 ′′ 2 f (x) = 12x - 72x
′′ f (x) = 0 when 2 12x - 72x = 0 ⇒ x(x - 6) = 0
So x = 0, x = 6.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.
 

Question 5

 
 
Find the points of inflexion of the function 2 f(x) = sin2x + x on the interval π 0 ≤ x ≤ -- 2 .
a) 0, π- 4   b) 0, π 2-
c) π- 6 , 5π- 6   d) π -- 12 , 5π --- 12

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
f(x) = sin2x+ x2 f′(x) = 2 cos2x + 2x ′′ f (x) = - 4sin 2x+ 2
f′′(x) = 0 when 1 4(-- sin2x) = 0 2 i.e when 1 sin2x = - 2
i.e
π 5π 2x = --,--- 6π 65π x = -- ,--. 12 12
 

Question 6

 
 
The graph below is of the function y = f(x).
PIC
Which of the following statements gives reasonable values for x which satisfy the conditions.
a) f′(x) < 0 when -1 < x < 3
f′′(x) < 0 when x > 1
f′(x) > 0 and f′′(x) = 0 when x = 1.		   b) ′ f (x) < 0 when -1 < x < 3
′′ f (x) < 0 when x > 1
′ f (x) > 0 and ′′ f (x) = 0 for no value of x.
c) ′ f (x) < 0 when x < -1 and x > 3
′′ f (x) > 0 when x < 1
′ f (x) > 0 and ′′ f (x) = 0 when x = 1.		   d) f′(x) < 0 when x < -1 and x > 3
f′′(x) > 0 when x < 1
f′(x) > 0 and f′′(x) = 0 for no value of x.

 

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

Question 7

 
 
The graph below is of the function y = f(x).
PIC
Which of the following statements gives reasonable values for x which satisfy the conditions.
a) f′(x) < 0 when x < 0 and 5 < x < 7
f′′(x) > 0 when 3 < x < 6
f′(x) > 0 and f′′(x) = 0 when x = 6.		   b) f′(x) < 0 when 1 < x < 5 and x > 7
f′′(x) > 0 when 3 < x < 6
f′(x) > 0 and f′′(x) = 0 when x = 6.
c) f′(x) < 0 when x < 1 and 5 < x < 7
f′′(x) > 0 when x < 3 and x > 6
f′(x) > 0 and f′′(x) = 0 when x = 3.		   d) ′ f (x) < 0 when 1 < x < 5 and 5 < x < 7
′′ f (x) > 0 when x < 3 and x > 6
′ f (x) > 0 and ′′ f (x) = 0 when x = 3.

 

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

Question 8

 
 
Find the value of a such that the function f(x) = xeax has a critical point at x = 3 . (Give your answer correct to 2 decimal places.)

 

Your answer is correct
We need f′(3) = e3a(3a + 1) = 0.
Not correct. You may try again.
Find f′(3), put it equal to zero, and solve for a.
 

Question 9

 
 
How many turning points are there on a graph of y = ex +sinx + 2x ?

 

Your answer is correct
dy- dx > 0 for all x , so there are no turning points.
Not correct. You may try again.
Note that dy- dx > 0 for all x .
 

Question 10

 
 
The graph of --2---- P (t) = 1+ e-t has exactly one point of inflexion. Find the value of P (t) at this point.

 

Your answer is correct
-t -t P′′(t) = 2e-(e----1) (1 + e-t)3 , and the only solution of P ′′(t) = 0 is t = 0 . So the point of inflexion must occur at t = 0 . Substituting t = 0 in the formula for P(t) gives P(0) = 1 .
Not correct. You may try again.
- t -t P ′′(t) = 2e-(e---t31) (1+ e )