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?

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- Aug 25th 2006, 07:33 PMstartingoverTangent 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? - Aug 25th 2006, 07:39 PMQuickdon't quote me
I can't help you because I don't know derivatives :o

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. - Aug 26th 2006, 06:32 AMGlaysher
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 $\displaystyle x = -1$