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

CalculusFunction  Steve, Monday, April 29, 2013 at 11:03am
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 = 1e^2
f" = e^x 1e^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^21
f(x) = e^x  1/2 (1+e^2)x^2 + 3x + e^21
You could do the same for trig functions, too, working with a translated graph:
f(x) = asin(b(x2)+c)
f' = abcos(b(x2)+c)
f" = ab^2 cos(b(x2)+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(x2)arccot(4/3))
Not sure on this one, but you see the method.
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