How do I go about finding all the values of x where the tangent line to y = f(x) is horizontal?
f(x) = x^3(2x^2 + x - 3)^2
I assume I would set f(x) and f '(x) equal to 0 and solve for x?
Quick is right the question is asking to find all the points where f '(x) = 0
$\displaystyle f(x) = x^3(2x^2 + x - 3)^2$
$\displaystyle u = x^3$ $\displaystyle v = (2x^2 + x - 3)^2$
Use product rule (with chain rule needed for differentiating v)
$\displaystyle u' = 3x^2$ $\displaystyle v' = 2(4x + 1)(2x^2 + x - 3)$
$\displaystyle f '(x) = 3x^2(2x^2 + x - 3)^2 + 2x^3(4x + 1)(2x^2 + x - 3)$
$\displaystyle 3x^2(2x^2 + x - 3)^2 + 2x^3(4x + 1)(2x^2 + x - 3) = 0$
Factorising
$\displaystyle x^2(2x^2 + x - 3)(3(2x^2 + x - 3) + 2x(4x + 1)) = 0$
$\displaystyle x^2(2x + 3)(x - 1)(14x^2 + 5x - 9) = 0$
$\displaystyle x^2 = 0$ gives $\displaystyle x = 0$
$\displaystyle 2x + 3 = 0$ gives $\displaystyle x = -\frac{3}{2}$
$\displaystyle x - 1 = 0$ gives $\displaystyle x = 1$
$\displaystyle 14x^2 + 5x - 9 = 0$ gives $\displaystyle (14x - 9)(x + 1) = 0$
Gives $\displaystyle x = \frac{9}{14}$ and $\displaystyle x = -1$