# Thread: Tangent Line to y=f(x)

1. ## Tangent Line to y=f(x)

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?

2. ## don't quote me

I can't help you because I don't know derivatives

however, as I recall from previous posts, you need to find the value of x when the f '(x)=0

finding when f(x)=0 does nothing.

3. Quick is right the question is asking to find all the points where f '(x) = 0

$f(x) = x^3(2x^2 + x - 3)^2$

$u = x^3$ $v = (2x^2 + x - 3)^2$

Use product rule (with chain rule needed for differentiating v)

$u' = 3x^2$ $v' = 2(4x + 1)(2x^2 + x - 3)$

$f '(x) = 3x^2(2x^2 + x - 3)^2 + 2x^3(4x + 1)(2x^2 + x - 3)$

$3x^2(2x^2 + x - 3)^2 + 2x^3(4x + 1)(2x^2 + x - 3) = 0$

Factorising

$x^2(2x^2 + x - 3)(3(2x^2 + x - 3) + 2x(4x + 1)) = 0$

$x^2(2x + 3)(x - 1)(14x^2 + 5x - 9) = 0$

$x^2 = 0$ gives $x = 0$

$2x + 3 = 0$ gives $x = -\frac{3}{2}$

$x - 1 = 0$ gives $x = 1$

$14x^2 + 5x - 9 = 0$ gives $(14x - 9)(x + 1) = 0$

Gives $x = \frac{9}{14}$ and $x = -1$