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
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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