# Find a, b, c, and d such that the cubic f(x) = ax3 + bx2 + cx + d satisfies the given conditions.?

0

Relative maximum: (2, 8)

Relative minimum: (4, 3)

Inflection point: (3, 5.5)

a =

b =

c =

d =

• Note that f ‘(x) = 3ax^2 + 2bx + c and f ”(x) = 6ax + 2b.

First of all, since we have a point of inflection at x = 3, we have

f ”(3) = 18a + 2b = 0 ==> b = -9a.

Moreover, since we have relative extrema at x = 2, 4, we have

f ‘(2) = 12a + 2(-9a) * 2 + c = 0 ==> c = 24a

So, f(x) = ax^3 – 9ax^2 + 24ax + d.

Now, we can solve for a and d using 2 of the 3 points:

(2, 8) ==> 8 = 20a + d

(4, 3) ==> 3 = 16a + d

Solving yields a = 5/4 and d = -17 = -68/4.

So, f(x) = (1/4) (5x^3 – 45x^2 + 120x – 68).

I hope this helps!

• all of us comprehend that: f(a million) = 0 f(2) = 9 Differentiating, we get: f'(a million) = 3 f'(2) = 18 So, we’ve the equipment: A + B + C + D = 0 8A + 4B + 2C + D = 9 3A + 2B + C = 3 12A + 4B + C = 18 fixing, we get: A = 3 B = -6 C = 6 D = -3 which factors us the polynomial function: ? f(x) = 3x³ – 6x² + 6x – 3 for which: f(a million) = 0 f(2) = 9

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