Calculus-Function

posted by on .

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.