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An inflection point of a function f(x) is the point where the curvature changes sign. When that happens f''(x) = 0.
Here an example of a function is required that has a point of inflection at (1, 4). Let the function be f(x).
f''(1) = 0
Let f''(x) = 6x - 6
f'(x) = 3x^2 - 6x + C
f(x) = x^3 - 3x^2 + C1*x + C2
f(1) = 4
=> 1^3 - 3*1 + C1*1 + C2 = 4
Let C2 = 0
=> C1 = 6
An example of a function that has an inflection point at (1,4) is f(x) = x^3 - 3x^2 + 6x
Hm if possible, could i have another expert answer? Its on my math review sheet and i wake to make sure :%
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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