prove that this equation has 1 real root

• June 15th 2008, 07:14 AM
Bryn
prove 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
• June 15th 2008, 07:19 AM
sean.1986
Make it of the form (x+a)(bx^2 + cx + d)

Then prove bx^2 + cx + d has no real roots.
• June 15th 2008, 07:25 AM
red_dog
$f(x)=x^3+8x-19, \ f'(x)=3x^2+8>0, \ \forall x\in\mathbf{R}$
So, f is strictly increasing.
$\lim_{x\to\infty}f(x)=\infty, \ \lim_{x\to\ -\infty}f(x)=-\infty$ 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.
• June 15th 2008, 08:07 AM
CaptainBlack
Quote:

Originally Posted by Bryn
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

See this thread. There is an entirly different proof there from any in this thread as yet.

RonL