y=x^3+8x-19

I know the derivative is 3x^2+8, but how does this tell me there is no turning point and therefore that the function is increasing has one root

thanks

Bryn

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- Jun 15th 2008, 07:14 AMBrynprove that this equation has 1 real root
y=x^3+8x-19

I know the derivative is 3x^2+8, but how does this tell me there is no turning point and therefore that the function is increasing has one root

thanks

Bryn - Jun 15th 2008, 07:19 AMsean.1986
Make it of the form (x+a)(bx^2 + cx + d)

Then prove bx^2 + cx + d has no real roots. - Jun 15th 2008, 07:25 AMred_dog
$\displaystyle f(x)=x^3+8x-19, \ f'(x)=3x^2+8>0, \ \forall x\in\mathbf{R}$

So, f is strictly increasing.

$\displaystyle \lim_{x\to\infty}f(x)=\infty, \ \lim_{x\to\ -\infty}f(x)=-\infty$ and f is continuous, so f has at least a real root. But f is injective (because is strictly increasing) and the root is unique. - Jun 15th 2008, 08:07 AMCaptainBlack
See this thread. There is an entirly different proof there from any in this thread as yet.

RonL