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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 $\alpha,\beta$

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  • $\begingroup$ Note the concavity of $f$ isn't a necessary assumption. You can just consider $f/g$ and its derivative. $\endgroup$ Sep 14, 2021 at 2:03

2 Answers 2

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This is a proof of the following

The slope of a concave upward function is negative at the first root and positive at the second, given that the roots in question are consecutive and the function has two or more roots.

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 $x > c$.

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  • $\begingroup$ What is $h(x)$? $\endgroup$ Sep 14, 2021 at 1:50
  • $\begingroup$ Just trying to use different notation not to confuse with the $f(x)$ in the problem statement. $\endgroup$
    – fwd
    Sep 14, 2021 at 1:50
  • $\begingroup$ My question wasn't clear. What I meant to ask was "What's the expression for $h(x)$?" Is $h(x)=f'(x)g(x)-g'(x)f(x)$ or something else? $\endgroup$ Sep 14, 2021 at 1:53
  • $\begingroup$ I am attempting to prove this: "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" $\endgroup$
    – fwd
    Sep 14, 2021 at 1:54
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    $\begingroup$ Ahh, I see. I'd recommend clarifying this in your post. $\endgroup$ Sep 14, 2021 at 1:56
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We can weaken the assumptions in the problem statement somewhat and still find an elementary proof. Consider the following alternative statement:

Let $f, g$ be differentiable functions of real numbers such that $f'(x)g(x) - g'(x)f(x) \neq 0$ for all $x$. Prove that $g(x)=0$ has at least one root between consecutive roots of $f(x)=0$.

To prove this, proceed as follows:

Let $\alpha < \beta$ be roots of $f(x)=0$.

For any root $x=c$ of $f(x)=0$, we have $$f'(c)g(c) = f'(c)g(c) - g'(c)f(c) \neq 0 \\ \implies g(c) \neq 0,$$ so in particular, we find that $\alpha, \beta$ are not roots of $g(x)=0$.

Now, towards a contradiction, suppose $g(x)\neq 0$ for all $\alpha < x < \beta$.

We have that $f,g$ are differentiable and $g(x) \neq 0$ for all $\alpha \leq x \leq \beta$. Therefore, the quotient $h(x) = f(x)/g(x)$ is continuous on $\alpha \leq x \leq \beta$, differentiable on $\alpha < x < \beta$, and satisfies $$h(\alpha) = f(\alpha)/g(\alpha) = 0 \\ h(\beta) = f(\beta)/g(\beta) = 0.$$

Therefore by the Mean Value Theorem (or more specifically Rolle's theorem), $$\frac{f'(x)g(x) - g'(x)f(x)}{(g(x))^2} = h'(x) = 0$$ has a root. However, this then implies the numerator has a zero, which contradicts the assumption $f'(x)g(x)-g'(x)f(x)\neq 0$ for all $x$. Therefore, our supposition that $g(x) \neq 0$ for all $\alpha < x < \beta$ must be incorrect, so $g(x)=0$ has a root between $\alpha, \beta$.

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