Find a cubic function, in the form below, that has a local maximum value of 3 at

Question

Find a cubic function, in the form below, that has a local maximum value of 3 at -4 and a local minimum value of 0 at 3. f(x) = ax3 + bx2 + cx + d f(x) =

Explanation / Answer

f'(x) = 3ax^2 + 2bx + c 0 = 3ax^2 + 2bx + c Using quadratic formula: x = [-2b +- sqrt(4b^2 - 4(3ac))]/6a These equal the relative extrema. But we have their values (0 and 3). So: [-2b - 2sqrt(b^2 - 3ac)]/6a = [-b - sqrt(b^2 - 3ac)/3a] = 0 => -b - sqrt(b^2 - 3ac) = 0 => -b = sqrt(b^2 -3ac) => b^2 = b^2 -3ac => ac = 0 a can't be zero, so c = 0. Also, [-2b - 2sqrt(b^2 - 3ac)]/6a = [-b + sqrt(b^2)]/3a = 3 => 9a = -b + sqrt(b^2) => 9a + b = sqrt(b^2) => 81a^2 + 18ab + b^2 = b^2 => 81a^2 + 18ab = 0 => a(81a + 18b) = 0 => 81a = -18b => a = (-2/9)b Since f(0) = 3, then we know d = 3. So we have f(x) = (-2/9)bx^3 + bx^2 + 3 If f(3) = -4, then -4 = (-2/9)b(3^3) + b(3^2) = 3 => -4 = -6b + 9b = 3 => -7 = 3b => b = (-7/3) So f(x) = (14/27)x^3 - (7/3)x^2 + 3

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