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Lesson: Finding a Function given Its Derivative

Worksheet • 25 Questions

Q1:

Determine 𝑓 ( 𝑑 ) if 𝑓 β€² β€² β€² ( 𝑑 ) = βˆ’ 4 √ 𝑑 + 5 𝑑 c o s .

  • A 𝑓 ( 𝑑 ) = βˆ’ 3 2 𝑑 1 0 5 βˆ’ 5 𝑑 + 𝑑 + 𝑑 + 7 2 s i n C D E 2
  • B 𝑓 ( 𝑑 ) = βˆ’ 3 2 𝑑 1 0 5 βˆ’ 5 𝑑 + 𝑑 + 7 2 s i n C D
  • C 𝑓 ( 𝑑 ) = βˆ’ 8 𝑑 3 + 5 𝑑 + 3 2 s i n C
  • D 𝑓 ( 𝑑 ) = βˆ’ 3 2 𝑑 1 0 5 βˆ’ 5 𝑑 + 𝑑 + 7 2 s i n C E 2
  • E 𝑓 ( 𝑑 ) = βˆ’ 1 6 𝑑 1 5 βˆ’ 5 𝑑 + 𝑑 + 5 2 c o s C D

Q2:

Find the equation of the curve passing though the point ( 0 , 1 ) given the gradient of the tangent at any point is 2 π‘₯ βˆ’ 3 . Then find the tangents at the points on the curve which intersect with the line 𝑦 = 5 .

  • Athe curve: 𝑦 = π‘₯ βˆ’ 3 π‘₯ + 1 2 , the tangents: 𝑦 βˆ’ 5 π‘₯ + 1 5 = 0 and 𝑦 + 5 π‘₯ = 0
  • Bthe curve: 𝑦 = π‘₯ βˆ’ 3 π‘₯ + 1 2 , the tangents: 𝑦 + 5 π‘₯ βˆ’ 2 5 = 0 and 𝑦 βˆ’ 5 π‘₯ βˆ’ 1 0 = 0
  • Cthe curve: 𝑦 = 2 π‘₯ βˆ’ 3 π‘₯ βˆ’ 1 2 , the tangents: βˆ’ 5 𝑦 βˆ’ π‘₯ + 2 9 = 0 and 5 𝑦 βˆ’ π‘₯ βˆ’ 2 6 = 0
  • Dthe curve: 𝑦 = π‘₯ βˆ’ 3 π‘₯ + 1 2 , the tangents: 5 𝑦 βˆ’ π‘₯ βˆ’ 2 1 = 0 and βˆ’ 5 𝑦 βˆ’ π‘₯ + 2 4 = 0

Q3:

The function 𝑦 = 𝑓 ( π‘₯ ) satisfies d d 2 2 𝑦 π‘₯ = π‘Ž π‘₯ + 𝑏 for some constants π‘Ž and 𝑏 . If the graph of 𝑓 has an inflection at ( 0 , 9 ) and a local minimum value at ( 3 , βˆ’ 9 ) , what is 𝑓 ( π‘₯ ) ? If the graph also has a local maximum, identify it.

  • A 𝑓 ( π‘₯ ) = 1 3 π‘₯ βˆ’ 9 π‘₯ + 9 3 , local maximum value: 𝑓 ( βˆ’ 3 ) = 2 7
  • B 𝑓 ( π‘₯ ) = π‘₯ + 1 8 π‘₯ + 9 3 , local maximum value: 𝑓 ( βˆ’ 3 ) = βˆ’ 5 4
  • C 𝑓 ( π‘₯ ) = 2 3 π‘₯ βˆ’ 3 6 π‘₯ + 9 3 , local maximum value: 𝑓 ( βˆ’ 3 ) = 1 0 8
  • D 𝑓 ( π‘₯ ) = 1 2 π‘₯ βˆ’ 1 8 π‘₯ + 9 3 , local maximum value: 𝑓 ( βˆ’ 3 ) = 5 4
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