Prove that when sum of bases (of a function $y$) is constant, the function $y$ will have a critical point, which can be found out by equating bases ratio to power ratio. ,i.e.

If $ y=f(x)^ng(x)^m$ and $f(x) + g(x) = constant$

then prove that $f(x)/g(x) = n/m$ gives us a root of $dy/dx$

How do I prove it?


i) $f(x)=(x-2)^5(3-x)^3$

Here Bases sum is $x-2+3-x =1$ (which is a constant) Therefore, one critical point of $f(x)$ can be given by $(x-2)/(3-x)=5/3$ upon solving this equation we get $x=21/8$ which is a root/zero $df(x)/dx$ and a critical point of $f(x)$.

ii) $f(x)=(\sin^6(x))(\cos^4(x))$

$f(x)$ can be rewritten as $f(x)=(\sin^2(x))^3(\cos^2(x))^2$ Again we find that bases sum is constant, i.e. $\sin^2(x)+\cos^2(x)=1$ Therefore, one critical point of $f(x)$ can be given by $\sin^2(x)/\cos^2(x)= 3/2$ upon solving this equation we get $x=0.88607712$ which is a root of $df(x)/dx$ and a critical point of $f(x)$.

  • $\begingroup$ Is $f(x)$ essentially a polynomial? $\endgroup$ – SchrodingersCat Dec 15 '15 at 13:44
  • $\begingroup$ It need not be. $\endgroup$ – Dimenein Dec 15 '15 at 13:46
  • $\begingroup$ Define exactly the set of functions you have in mind, and what you mean by the "bases" of such an $f$. $\endgroup$ – Christian Blatter Dec 15 '15 at 14:21

All you need to prove this is some basic substitutions. I'll just do it with two bases raised to powers but it could be easily extended to more.

So based on what you've given, we have

$ y=f(x)^ng(x)^m $

$ f(x) + g(x) = c $ where c is a constant, and

$ f(x)/g(x) = n/m $

from these we get

$ \dfrac{dy}{dx} = nf(x)^{n-1}g(x)^mf'(x) + mf(x)^ng(x)^{m-1}g'(x) $ which must equal zero

$ n=mf(x)/g(x) $

$ g(x)=c-f(x) $ which gets us

$ g'(x)=-f'(x) $

when we plug that it we get

$mf(x)/g(x)*f(x)^{n-1}g(x)^mf'(x) + mf(x)^ng(x)^{m-1}(-f'(x))$

which then gets us

$mf(x)^ng(x)^{m-1}f'(x) - mf(x)^ng(x)^{m-1}f'(x) = 0$

so $\dfrac{dy}{dx} = 0$ and we get a critical point

  • $\begingroup$ In the second example , $f(x)=(\sin^6(x))(\cos^4(x))$ if we consider the sum of bases directly $(\sin^6(x))+(\cos^4(x))$ then it would give us tanx=6/4 which is not root/zero of df(x)/dx. $\endgroup$ – Dimenein Dec 16 '15 at 13:41
  • $\begingroup$ @Vinayak You're right, I had made a mistake. I've fixed it and the math works out. $\endgroup$ – rtpax Dec 16 '15 at 17:59

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