MATH 105 Lecture Notes - Lecture 10: Glutamine, Antiderivative, Muumuu

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bronzebear210

18 Feb 2020

School

UBC

Department

Mathematics

Course

MATH 105

Professor

ISSA, AHMAD ISSA KHALID

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MATH 105 Lecture Notes - Lecture 9: Riemann Sum

MATH 105 Lecture Notes - Lecture 10: Glutamine, Antiderivative, Muumuu

MATH 105 Lecture Notes - Lecture 11: International Fund For Agricultural Development, Adze

Document Summary

Eg suppose that fothdn 3 and fhdn to. An antiderivative of gln is my function gcn such that g gg. Eg find an antiderivative of a gcn 2. So gcn n is an anti derivative of gcn i danger. Q w n ttoo also an antiderivative of lu retlool. I c where c is a constant are all the antiderivatives of 2n. We denote the general antiderivative by antidifferentiatewithrespectto u. 2ndn n"t c where c is a constant indefiniteintegralof 2n f t du. J n dn n last day we used riemann turns to show fo n du. Can compute signed areas using antiderivatives why we"ll see today before getting thee we need. Aln is the area function of fia with left endpoint a. Could use riemanndms but regren isjust a trapezium y a f. Itt h f f n h n 2. What isthe area under y fct7 between f 2 and f 4.

Related Questions

Hello, I am having an extremely diffucult time understanding trig subs in Calc2. Can someone please fully expalin to me why the limits of integration change from 3sqrt3/2 to pi/3?....and than change once again from 1 to 1/2. I know you can plug the number somewhere to see if they change, but I'm not sure where to do this...or even WHY to do this? I still don't understand why we change limits of integration in other calc problems like u-sub. Can you please explain it like you're explaining it to someone who has no math skills. I really appreiciate a clear explanation without too much math terminology.

God bless and thank you!

Find the derivative of the function and evaluate thederivative at the given x-value.

f(x) = 2x² at x = 1

A) f' (x) = 4x; f' (1) = 2
B) f ' (x) = 2x; f ' (1) = 2
C) f' (x) = 4x²; f' (1) = 4
D) f' (x) = 4x; f' (1) = 4

f(x) = 6x + 2; [-1, 2]

A) Absolute maximum: 12, absolute minimum: -6
B) Absolute maximum: -1, absolute minimum: 2
C) There are no absolute extrema.
D) Absolute maximum: 14, absolute minimum: -4

f(x) = (-5x + 7)⁴

A) f '(x) = -5(-5x + 7)³
B) f '(x) = -20(-5x + 7)⁴
C) f '(x) = 4(-5x + 7)³
D) f '(x) = -20(-5x + 7)³

A)
B)
C)
D)

A company estimates that the daily revenue (in dollars) from thesale of x cookies is given by

R(x) = 885 + 0.02x + 0.0003x²

Currently, the company sells 900 cookies perday.

Use marginal revenue to estimate the increase in revenueif the company increases sales by one cookie per day.

A) $92.00
B) $0.56
C) $56.00
D) $0.92

f(x) = -3 - 7x; [-3, 1]

A) Absolute maximum: -10, absolute minimum: -24
B) Absolute maximum: 18, absolute minimum: -10
C) There are no absolute extrema
D) Absolute maximum: 24, absolute minimum: -4

A)
B)
C)
D)

s(x) = -x² - 20x - 19

A) Relative maximum at ( -20, -19)
B) Relative maximum at ( 10, 81)
C) Relative maximum at ( -10, 81)
D) Relative minimum at ( 20, -19)

Differentiate.

f(x) = (5x + 4)²

A) f'(x) = 10(5x + 4)²
B) f'(x) = 5(5x + 4)
C) f'(x) = 2(5x + 4)
D) f'(x) = 10(5x + 4)

10.

Find the absolute maximum and absolute minimum values ofthe function, if they exist, over the indicated interval. When nointerval is specified, use the real line (-?, ?).

f(x) = -21; [ -7, 7]

A) Absolute maximum: 21, absolute minimum: 0
B) Absolute maximum: 21, absolute minimum: -21
C) Absolute maximum: -21, absolute minimum: -21
D) There are no absolute extrema.

11.

A)
B)
C)
D)

12.

f(x) = -6x² - 2x - 7

A) Relative maximum at
B) Relative maximum at
C) Relative maximum at
D) Relative maximum at

13.

f(x) = 0.2x² - 2.4x + 5.9

A) Relative minimum at ( 6, -1.3)
B) Relative maximum at ( 6, -1.3)
C) Relative minimum at ( -6, 27.5)
D) Relative minimum at ( 6, 0)

14.

A grocery store estimates that the weekly profit (in dollars)from the production and sale of x cases of soup is given by