finding functions | Mathematics | Page 1

willow smelt Dec 31, 2022, 4:34 PM

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crude stirrup Dec 31, 2022, 4:43 PM

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willow smelt

We have:
f(x) = ax^3 + bx^2 + cx + d
f'(x) = 3ax^2 + 2bx + c
On the other hand, we must have f'(x) = 0 at x = -3, x = 2. So:
3ax^2 + 2bx + c = 3a(x - 2)(x + 3)
We expand it:
3ax^2 + 2bx + c = 3ax^2 + 3ax - 18a
2bx + c = 3ax - 18a
So, from here we get b = (3/2)a, c = -18a. Thus:
f(x) = a(x^3 + (3/2)x^2 - 18x) + d
Finally, to find the last two constants, substitute the coordinates of the given points into this equation. You will get a linear system with two equations in two unknowns a, d. Solve it to get the answer.

willow smelt Dec 31, 2022, 4:49 PM

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ok thanks

crude stirrup Dec 31, 2022, 4:50 PM

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You're welcome!

willow smelt Dec 31, 2022, 4:53 PM

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how would you solve to find a and d

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