If $y=f(x)$ is a concave upward function and $y=g(x)$ is a function
such that $f'(x)g(x)-g'(x)f(x)=x^4+2x^2+10$ then prove that $g(x)$ has at least one root between consecutive roots of $f(x)=0$
$f'(x)g(x)-g'(x)f(x)=(x^2+1)^2+9\implies f'(x)g(x)-g'(x)f(x)>0$
If $\alpha,\beta$ are consecutive roots of $f(x)=0$, $f'(\alpha)g(\alpha)>0$ and $f'(\beta)g(\beta)>0$
Now if we prove that: "The slope of a concave upward function is positive at the first root and negative at the second, given that the roots in question are consecutive and the function has two or more roots", we will be able to reach the result.
When I draw a graph, the above statement seems obvious, but I don't know how to prove it mathematically
Because $g(\alpha)g(\beta)<0 \implies g(x)$ has at least one root in $\alpha,\beta$
Best Answer
This is a proof of the following
Let $h(x)$ be differentiable, concave up and such that $h(\alpha) = h(\beta) = 0$ (where $\alpha, \beta$ are consecutive roots), then $h'(x)$ is increasing. $h(x)$ attains a minimum on $[\alpha, \beta]$, and it is quite easy to see that this minimum must occur in $(\alpha, \beta)$ if $h$ is not $0$ on $[\alpha, \beta]$. Thus $h'(c) = 0$ for some $c \in (\alpha, \beta)$. Since $h'(x)$ is increasing, $h'(x) \le 0$ for all $x < c$ and $h'(x) \ge 0$ for $x > c$.