which I cannot do :P
Well, I just dont remember the best method to answer it
"$\displaystyle f(x)=x^3+8x-19$
show that the equation $\displaystyle f(x)=0$ has only one real root."
Thanks a lot
$\displaystyle f'(x) = 3x^2 + 8 > 0$
Thus f(x) is an increasing function. Now by intermediate value theorem $\displaystyle \exists c \in (1,2): f(c) = 0$.
Since the function is increasing, $\displaystyle \forall x > c, f(x) > 0; \forall x < c, f(x) < 0 $.
Thus $\displaystyle f(x) \neq 0, x \neq c $ and hence f(x) has only one real root.
By Descartes rule of signs it has at most 1 positive real root.
Also changing the sign of all odd power terms and applying Descartes rule of signs agrain we see that it has at most 0 negative real roots.
But this is a cubic and so has at leat one real root, and so it must have exactly one real root and its positive.
RonL