Show that the function $x^4 – 3x^2 + x-10$ cannot have a root inside $(0,2)$. Show that the function $x^4 – 3x^2 + x-10$ cannot have a root inside $(0,2)$.
Please note that roots of $f'(x)$ cannot be found using a calculator. Attempted the question by calculating $f'(x)$ and assuming that at least one root of $f(x)$ exists in $(0,2)$. However had difficulty locating the points of maxima/minima, which cannot be done without finding the roots to $f'(x)$. Think there's another way to do it. Any suggestions?
 A: Note that the function $g(x) = x^4-3x^2$ achieves maximum modulus in $[0,2]$ at $x=2$, with $|g(4)|=4$. Also note that in $(0,2)$ we have
$$|x^4-3x^2+x-10|>10-|x|-|g(x)| > 8-|g(x)| \geq 4.$$
A: You can show that in the domain $(0,2)$, 
$$ x^4 - 3x^2 + x - 10 < -10 + 3x.$$
This is equivalent to
$$x (x-2) (x+1)^2 < 0. $$
Since $-10 + 3x < 0 $ in the domain $(0,2)$, it follows that the function is never 0 in the domain.

The RHS of the inequality is obtained by finding the linear function which satisfies the values at the end points. This helps us by ensuring that we know 2 of the roots.
A: Calvin's method looks great [+1].  If you really want other ways, you could try
If $0 < x < 2$, then $x^3 < 8$ and $3x^2 + 10 > 2 \sqrt{30} x$ by AM-GM.
So $x^4 + x = x(x^3+1) < 9 x < 2 \sqrt{30} x < 3x^2 + 10$
A: Each term of a polynomial function $f:x\mapsto \sum_{k=0}^na_kx^k$ is monotonic in $x$ over any interval $[a,b]$ with $0\leqslant a\leqslant b$. Therefore $$f(x)\le \sum_{k=0}^n a_kc_k^k\quad(a\leqslant x\leqslant b),$$where $c_k=a$ if $a_k\leqslant 0$ and $c_k=b$ if $a_k>0.$ Now you can apply this to your example by checking this upper bound in each of (say) the intervals $[0,\sqrt2]$, $[\sqrt2,\sqrt3],$ and $[\sqrt3,2]$, which is easy to calculate.
