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Calculus-Function

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sketch the graph of a function in neighborhood x=2 that satisfies these conditions

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 - ,

    There 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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