<p>Let f(x) be the function defined by f(x) = k + 12x +3x^2-2x^3, where k is constant.</p>
<p>If the relative maximum of f is 4, what is the value of k?
the answer is -16, but I dont know how to get it at all...
please help thanks</p>
<p>f(x) = k + 12x +3x^2-2x^3
f'(x) = 12 + 6x + 6x^2</p>
<p>To find max & min points equate f'(x) to 0 (critical points):
f'(x) = 12 + 6x + 6x^2 = 0
x = 2 or x = -1</p>
<p>Since f'(x) changes from + to - at x=2, the relative max occurs at x=2. You also know that y = 4. Therefore, plugging this into the original equation gives:</p>
<p>f(x) = k + 12x +3x^2-2x^3
4 = k + 12(2) + 3(2^2) - 2(2^3)
4 = k + 20
k = -16</p>
<p>:)</p>
<p>aggh i knew how to do that snap i read the question wrong, i thought it meant when x=4 it had max...thanks!</p>