1. A Simple Question

which I cannot do :P

Well, I just dont remember the best method to answer it

" $f(x)=x^3+8x-19$

show that the equation $f(x)=0$ has only one real root."

Thanks a lot

2. Originally Posted by Silver
which I cannot do :P

Well, I just dont remember the best method to answer it

" $f(x)=x^3+8x-19$

show that the equation $f(x)=0$ has only one real root."

Thanks a lot
$f'(x) = 3x^2 + 8 > 0$

Thus f(x) is an increasing function. Now by intermediate value theorem $\exists c \in (1,2): f(c) = 0$.

Since the function is increasing, $\forall x > c, f(x) > 0; \forall x < c, f(x) < 0$.

Thus $f(x) \neq 0, x \neq c$ and hence f(x) has only one real root.

3. Originally Posted by Silver
which I cannot do :P

Well, I just dont remember the best method to answer it

" $f(x)=x^3+8x-19$

show that the equation $f(x)=0$ has only one real root."

Thanks a lot
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