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Quiz 6: One variable optimisation

Question

Question 1

Find the relative growth rate of the function $P (t) = 10e0.3t$ .

Your answer is correct

The relative growth rate is $′ P-(t)- P (t)$ .

Not correct. You may try again.

Remember that the relative growth rate is $P-′(t) P(t)$ .

Question 2

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 inflection.

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

False.
False. There is a local maximum value of 7, but it is not the absolute maximum.
True.
False. There is a local maximum at x = 0.
True.

Question 3

Find the absolute maximum of the function $x+ 3 f(x) = x2 +-x-- 2$ on the interval -2 ≤ x ≤ 1.

a)	$- 2- 18$	b)	-10
c)	-1	d)	-5

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Your answer is correct.

$f′(x) = - (x-+-5)(x-+-1) (x + 2)(x - 1)$
So the critical value of f on the given interval is at x = -1.
Now f(-1) = -1 and f(-2) = f(1) = -∞ so the absolute maximum of the function is -1.

Not correct. Choice (d) is false.

Question 4

Find the absolute minimum of the function $t2 + 2t f(t) =---t-- e$ on the interval -2 ≤ t ≤ 2.

a)	$√ - 2--2√--2 e- 2$	b)
c)	$2 + 2√2 ---√2-- e$	d)	-

Your answer is correct.

$f′(x) = - 2--t2 et$
Critical points are at t = ± √ -
2

. There is a minimum at t = - √ -
2

.
$√- f(- √2) = 2--2√-2-≈ - 3.41 e- 2$
f(2) = 1.08, f(-2) = 0
therefore the absolute minimum is $√ - 2--2√--2 e- 2$ 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 5

Find the points of inflection of the function $f(x) = x4 - 12x3 + 6x- 9$ on the interval -2 ≤ x ≤ 10.

a)	x = 0, 6	b)	x = 0, -6
c)	x = ±	d)	x = ±

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 6

Find the points of inflection 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 7

The graph below is of the function y = f(x).

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 8

The graph below is of the function y = f(x).

Which of the following statements gives reasonable values for x which satisfy the conditions.

a)	$′ f (x) < 0$ when x < 1 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.