The tangent line, where a function increases and decreases, stationary points and the extremes on a closed interval, the second and higher derivatives, and inflection points. Set for OMPT-A, OMPT-B, OMPT-D and OMPT-F.
On which interval is f(x)=x3−9x2+15x+2 decreasing?
Answer:
2
Give the third derivative f′′′(x) of f(x)=3x5−2x4+5x3−8x.
Answer:
3
f(x)=4x3+12x2−36x. Give the x-coordinate of the local MAXIMUM.
Answer:
4
Give the x-coordinate of the inflection point of f(x)=x3−12x2+36x+4.
Answer:
5
Give the equation of the tangent line to the graph of f(x)=−x2−x+1 at the point where x=−2, in the form y=mx+c.
Answer:
6
Give the local maximum value of f(x)=x3+9x2+15x−3.
Answer:
7
Give the maximum value of f(x)=x3−12x2+36x−5 on the interval [0,9].
Answer:
8
f(x)=2x3+6x2−18x. Give the x-coordinate of the local MINIMUM.
Answer:
9
Give f′′(3) for f(x)=x3−6x2−2.
Answer:
10
For f(x)=x3−6x2−2: at x=5, is f increasing or decreasing, and is the graph getting steeper or less steep? (Answer, for example: increasing and getting steeper.)
Answer:
11
On which interval is f(x)=x3−9x2+15x−1 decreasing?
Answer:
12
Give the third derivative f′′′(x) of f(x)=2x5−x4−3x3+6x.
Answer:
13
f(x)=x3−6x2−63x. Give the x-coordinate of the local MAXIMUM.
Answer:
14
Give the x-coordinate of the inflection point of f(x)=x3−15x2+63x+5.
Answer:
15
Give the equation of the tangent line to the graph of f(x)=x3−4x2−2x+1 at the point where x=3, in the form y=mx+c.
Answer:
16
Give the local maximum value of f(x)=x3−6x2+9x+3.
Answer:
17
Give the minimum value of f(x)=x3+12x2+45x−2 on the interval [−7,−1].
Answer:
18
f(x)=4x3−24x2−252x. Give the x-coordinate of the local MINIMUM.
Answer:
19
Give f′′(−3) for f(x)=x3−3x2+5.
Answer:
20
For f(x)=x3−27x−1: at x=−2, is f increasing or decreasing, and is the graph getting steeper or less steep? (Answer, for example: increasing and getting steeper.)
Answer:
21
On which interval is f(x)=x3−6x2−2 decreasing?
Answer:
22
Give the third derivative f′′′(x) of f(x)=2x5−3x4+2x3+7x.
Answer:
23
f(x)=2x3+9x2−24x. Give the x-coordinate of the local MAXIMUM.
Answer:
24
Give the x-coordinate of the inflection point of f(x)=x3+6x2−2.
Answer:
25
Give the equation of the tangent line to the graph of f(x)=x3−x2+6x−5 at the point where x=−1, in the form y=mx+c.
Answer:
26
Give the local maximum value of f(x)=x3−9x2+24x−1.
Answer:
27
Give the maximum value of f(x)=x3−15x2+48x+1 on the interval [0,12].
Answer:
28
f(x)=4x3+6x2−504x. Give the x-coordinate of the local MINIMUM.
Answer:
29
Give f′′(1) for f(x)=x3−9x2+15x+4.
Answer:
30
For f(x)=x3−3x2−24x+2: at x=5, is f increasing or decreasing, and is the graph getting steeper or less steep? (Answer, for example: increasing and getting steeper.)
Answer:
Answers
Every answer below was re-derived independently before this page was built.
1
Answer:[1,5]
f′(x)=3x2−18x+15=3(x−1)(x−5), a parabola opening upwards with zeros at 1 and 5, so f′(x)≤0 between them: f decreases on [1,5].
Common mistakes, and the answer each one gives:
Gave where f increases instead → (−∞,1]∪[5,∞)
Read the zeros of f′ off the factors with the signs swapped → [−5,−1]
2
Answer:180x2−48x+30
Differentiate three times: f′(x)=15x4−8x3+15x2−8,f′′(x)=60x3−24x2+30x, and f′′′(x)=180x2−48x+30.
Common mistakes, and the answer each one gives:
Stopped at the second derivative → 60x3−24x2+30x
Differentiated once too often → 360x−48
3
Answer:−3
f′(x)=12x2+24x−36=12(x+3)(x−1), which is zero at x=−3 and x=1.f′′(x)=24x+24, and f′′(−3)=−48<0, so x=−3 is the local maximum.
Common mistakes, and the answer each one gives:
Gave the local MINIMUM instead → 1
Gave the y-value rather than the x-coordinate → 108
4
Answer:4
At an inflection point f′′ changes sign. f′′(x)=6x−24, which is zero at x=4 and changes sign there, from negative to positive.
Common mistakes, and the answer each one gives:
Solved f′(x)=0 and gave the smaller root → 2
Solved f′′(x)=0 with the sign turned round → −4
5
Answer:y=3x+5
The slope is the derivative there: f′(x)=−2x−1, so m=f′(−2)=3. The point is (−2,f(−2))=(−2,−1). So y=3(x+2)−1, which is y=3x+5.
Common mistakes, and the answer each one gives:
Used the y-valuef(−2) as the slope → y=−x−3
Wrote y=f′(a)x+f(a), leaving out the shift to x=a → y=3x−1
6
Answer:22
f′(x)=3(x+5)(x+1) is zero at x=−5 and x=−1.f′′(x)=6x+18 is negative at x=−5, so that is the maximum, and its value is f(−5)=22.
Common mistakes, and the answer each one gives:
Gave the x-coordinatex=−5 instead of the value → −5
Gave the local minimum value instead → −10
7
Answer:76
On a closed interval the extremes are at a stationary point or at an end. f′(x)=3(x−2)(x−6) is zero at x=2 and x=6; compare f(0)=−5,f(2)=27,f(6)=−5 and f(9)=76. The maximum is 76, at the end x=9.
Common mistakes, and the answer each one gives:
Gave the local maximum, without checking the ends → 27
Gave the x-coordinatex=9 instead of the value → 9
8
Answer:1
f′(x)=6x2+12x−18=6(x+3)(x−1), which is zero at x=−3 and x=1.f′′(x)=12x+12, and f′′(1)=24>0, so x=1 is the local minimum.
Common mistakes, and the answer each one gives:
Gave the local MAXIMUM instead → −3
Gave the y-value rather than the x-coordinate → −10
9
Answer:6
f′(x)=3x2−12x and f′′(x)=6x−12, so f′′(3)=6.
Common mistakes, and the answer each one gives:
Gave f′(3) instead → −9
Gave f(3) instead → −29
10
Answer:increasing and getting steeper
f′(5)=15 is positive, so f is increasing. f′′(5)=18: the slope itself is rising, and a positive slope that is rising gets steeper.
Common mistakes, and the answer each one gives:
Read the direction from the sign of f(5) instead of f′(5) → decreasing and getting steeper
11
Answer:[1,5]
f′(x)=3x2−18x+15=3(x−1)(x−5), a parabola opening upwards with zeros at 1 and 5, so f′(x)≤0 between them: f decreases on [1,5].
Common mistakes, and the answer each one gives:
Gave where f increases instead → (−∞,1]∪[5,∞)
Read the zeros of f′ off the factors with the signs swapped → [−5,−1]
12
Answer:120x2−24x−18
Differentiate three times: f′(x)=10x4−4x3−9x2+6,f′′(x)=40x3−12x2−18x, and f′′′(x)=120x2−24x−18.
Common mistakes, and the answer each one gives:
Stopped at the second derivative → 40x3−12x2−18x
Differentiated once too often → 240x−24
13
Answer:−3
f′(x)=3x2−12x−63=3(x+3)(x−7), which is zero at x=−3 and x=7.f′′(x)=6x−12, and f′′(−3)=−30<0, so x=−3 is the local maximum.
Common mistakes, and the answer each one gives:
Gave the local MINIMUM instead → 7
Gave the y-value rather than the x-coordinate → 108
14
Answer:5
At an inflection point f′′ changes sign. f′′(x)=6x−30, which is zero at x=5 and changes sign there, from negative to positive.
Common mistakes, and the answer each one gives:
Solved f′(x)=0 and gave the smaller root → 3
Solved f′′(x)=0 with the sign turned round → −5
15
Answer:y=x−17
The slope is the derivative there: f′(x)=3x2−8x−2, so m=f′(3)=1. The point is (3,f(3))=(3,−14). So y=(x−3)−14, which is y=x−17.
Common mistakes, and the answer each one gives:
Used the y-valuef(3) as the slope → y=−14x+28
Wrote y=f′(a)x+f(a), leaving out the shift to x=a → y=x−14
16
Answer:7
f′(x)=3(x−1)(x−3) is zero at x=1 and x=3.f′′(x)=6x−12 is negative at x=1, so that is the maximum, and its value is f(1)=7.
Common mistakes, and the answer each one gives:
Gave the x-coordinatex=1 instead of the value → 1
Gave the local minimum value instead → 3
17
Answer:−72
On a closed interval the extremes are at a stationary point or at an end. f′(x)=3(x+5)(x+3) is zero at x=−5 and x=−3; compare f(−7)=−72,f(−5)=−52,f(−3)=−56 and f(−1)=−36. The minimum is −72, at the end x=−7.
Common mistakes, and the answer each one gives:
Gave the local minimum, without checking the ends → −56
Gave the x-coordinatex=−7 instead of the value → −7
18
Answer:7
f′(x)=12x2−48x−252=12(x+3)(x−7), which is zero at x=−3 and x=7.f′′(x)=24x−48, and f′′(7)=120>0, so x=7 is the local minimum.
Common mistakes, and the answer each one gives:
Gave the local MAXIMUM instead → −3
Gave the y-value rather than the x-coordinate → −1568
19
Answer:−24
f′(x)=3x2−6x and f′′(x)=6x−6, so f′′(−3)=−24.
Common mistakes, and the answer each one gives:
Gave f′(−3) instead → 45
Gave f(−3) instead → −49
20
Answer:decreasing and getting steeper
f′(−2)=−15 is negative, so f is decreasing. f′′(−2)=−12: the slope itself is falling, and a negative slope that is falling gets steeper.
Common mistakes, and the answer each one gives:
Took f′′>0 to mean steeper, whatever the direction → decreasing and getting less steep
Read the direction from the sign of f(−2) instead of f′(−2) → increasing and getting steeper
21
Answer:[0,4]
f′(x)=3x2−12x=3x(x−4), a parabola opening upwards with zeros at 0 and 4, so f′(x)≤0 between them: f decreases on [0,4].
Common mistakes, and the answer each one gives:
Gave where f increases instead → (−∞,0]∪[4,∞)
Read the zeros of f′ off the factors with the signs swapped → [−4,0]
22
Answer:120x2−72x+12
Differentiate three times: f′(x)=10x4−12x3+6x2+7,f′′(x)=40x3−36x2+12x, and f′′′(x)=120x2−72x+12.
Common mistakes, and the answer each one gives:
Stopped at the second derivative → 40x3−36x2+12x
Differentiated once too often → 240x−72
23
Answer:−4
f′(x)=6x2+18x−24=6(x+4)(x−1), which is zero at x=−4 and x=1.f′′(x)=12x+18, and f′′(−4)=−30<0, so x=−4 is the local maximum.
Common mistakes, and the answer each one gives:
Gave the local MINIMUM instead → 1
Gave the y-value rather than the x-coordinate → 112
24
Answer:−2
At an inflection point f′′ changes sign. f′′(x)=6x+12, which is zero at x=−2 and changes sign there, from negative to positive.
Common mistakes, and the answer each one gives:
Solved f′(x)=0 and gave the smaller root → −4
Solved f′′(x)=0 with the sign turned round → 2
25
Answer:y=11x−2
The slope is the derivative there: f′(x)=3x2−2x+6, so m=f′(−1)=11. The point is (−1,f(−1))=(−1,−13). So y=11(x+1)−13, which is y=11x−2.
Common mistakes, and the answer each one gives:
Used the y-valuef(−1) as the slope → y=−13x−26
Wrote y=f′(a)x+f(a), leaving out the shift to x=a → y=11x−13
26
Answer:19
f′(x)=3(x−2)(x−4) is zero at x=2 and x=4.f′′(x)=6x−18 is negative at x=2, so that is the maximum, and its value is f(2)=19.
Common mistakes, and the answer each one gives:
Gave the x-coordinatex=2 instead of the value → 2
Gave the local minimum value instead → 15
27
Answer:145
On a closed interval the extremes are at a stationary point or at an end. f′(x)=3(x−2)(x−8) is zero at x=2 and x=8; compare f(0)=1,f(2)=45,f(8)=−63 and f(12)=145. The maximum is 145, at the end x=12.
Common mistakes, and the answer each one gives:
Gave the local maximum, without checking the ends → 45
Gave the x-coordinatex=12 instead of the value → 12
28
Answer:6
f′(x)=12x2+12x−504=12(x+7)(x−6), which is zero at x=−7 and x=6.f′′(x)=24x+12, and f′′(6)=156>0, so x=6 is the local minimum.
Common mistakes, and the answer each one gives:
Gave the local MAXIMUM instead → −7
Gave the y-value rather than the x-coordinate → −1944
29
Answer:−12
f′(x)=3x2−18x+15 and f′′(x)=6x−18, so f′′(1)=−12.
Common mistakes, and the answer each one gives:
Gave f′(1) instead → 0
Gave f(1) instead → 11
30
Answer:increasing and getting steeper
f′(5)=21 is positive, so f is increasing. f′′(5)=24: the slope itself is rising, and a positive slope that is rising gets steeper.
Common mistakes, and the answer each one gives:
Read the direction from the sign of f(5) instead of f′(5) → decreasing and getting steeper