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
$\displaystyle f(x)=x^3+8x-19, \ f'(x)=3x^2+8>0, \ \forall x\in\mathbf{R}$
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.
See this thread. There is an entirly different proof there from any in this thread as yet.
RonL