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January 27, 2015

January 27, 2015

Posted by **Liliana** on Monday, April 29, 2013 at 12:22am.

f(2)=3

f'(2)=2

f"(2)=-1

you can find a specific function that satisfies all of the given conditions

i know the first derivative gives you critical points, and second derivative gives you points of inflection/concavity but how would you do this. i have no clue where to start. thankyou

- Calculus-Function -
**Steve**, Monday, April 29, 2013 at 11:03amThere are lots of possibilities here. The easiest is probably a simple parabola, with constant f" = -1

f' = -x + c

since f'(2) = 2,

2 = -2 + c

c=4

f' = -x+4

f = -1/2 x^2 + 4x + c

since f(2) = 3,

3 = -2 + 8 + c

c = -3

f(x) = -1/2 x^2 + 4x - 3

Or, you could try an exponental

f" = e^x+c

-1 = e^2+c

c = -1-e^2

f" = e^x -1-e^2

f' = e^x - (1+e^2)x + c

2 = e^2 - (1+e^2) + c

c = 3

f' = e^x - (1+e^2)x + 3

f = e^x - 1/2 (1+e^2)x^2 + 3x + c

3 = e^2 - 2(1+e^2) + 6 + c

c = e^2-1

f(x) = e^x - 1/2 (1+e^2)x^2 + 3x + e^2-1

You could do the same for trig functions, too, working with a translated graph:

f(x) = asin(b(x-2)+c)

f' = abcos(b(x-2)+c)

f" = -ab^2 cos(b(x-2)+c)

3 = asin(c)

2 = abcos(c)

-1 = ab^2 cos(c)

b = -1/2

2/3 = -1/2 cot(c)

so, cot(c) = -4/3

so, 3 = a (-4/5)

c = -15/4

f(x) = -15/4 sin(-1/2(x-2)-arccot(4/3))

Not sure on this one, but you see the method.

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