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, 08:33 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, 08:39 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, 07:32 AMGlaysher
Quick is right the question is asking to find all the points where f '(x) = 0

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

Factorising

gives

gives

gives

gives

Gives and