Determine the value or values of \(x\) for which the tangent to \(f\) is horizontal by first finding the derivative of \(f\) with respect to \(x\) then solving \(f'(x) = 0\) for \(x\).
PART 1.
\(f(x) = 2x^3 - 48x^2 + 360x\)
\(f'(x) = 6x^2 - 96x + 360\)
NOTE: for this problem, you should use the definition of derivative, \(f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\) or the equivalent form \(f'(x) = \lim_{z \to x} \frac{f(z) - f(x)}{z - x}\)
PART 2.
The tangent to the curve is horizontal at: \(x = 800, x = 864\)
NOTE: Type answer in form \(x = c\). Separate multiple answers with a comma, such as \(x = 1, x = -1\)
Please note that you may use the rules for finding derivatives for this type of question on the
