The function has an inflection point when the 2nd
derivative is cancelling out.
We'll have to choose a
function of 3rd order, to be differentiated twice.
f(x) =
ax^3 + bx^2 + cx + d
We'll differentiate with respect to
x:
f'(x) = 3ax^2 + 2bx +
c
We'll differentiate with respect to x
again:
f"(x) = 6ax + 2b
We'll
put f"(x) = 0 for x = 1
6a + 2b =
0
3a + b = 0
b =
-3a
We'll calculate f'(1) = 3a + 2b +
c
f'(1) = b + c
f(1) = a + b +
c + d
But f(1) = 4
a + b + c +
d = 4
To determine the function f(x), it would be necessary
to provide another constraint concerning the 1st derivative, otherwise, the coefficients
cannot be determined under the
circumstances.
But, you have to remember that
the function is a polynomial of 3rd order, at least: f(x) = ax^3 + bx^2 + cx +
d.
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