Without using calculus, try guessing the maximum value of

Question

kayders1997 · Answer

Of?

rational · Answer

$$\large f(x)=\frac{a}{\sqrt{b+\left(x-\frac{c^2}{x}\right)^2}}$$

rational · Answer

$a,b,c$ are constants

priyar · Answer

at x =0 right?

rational · Answer

at x = 0, the bottom expression becomes infinity !

priyar · Answer

oh! i didn't see that x as denominator .. sry

rational · Answer

Btw, the given function may represent the current through a driven series RLC circuit.

maximum value represents the resonance condition..

rational · Answer

But it seems you have the right idea @priyar

parthkohli · Answer

$x = c$?

parthkohli · Answer

I mean that squared expression can take $0$ at the least which is at $x = c$

rational · Answer

Yes!

priyar · Answer

yes i think u r right! coz its  max. when the bracket part is  zero

priyar · Answer

nice Q @rational !

parthkohli · Answer

I guess $\pm c$ would be better. But yeah, same idea.

phi · Answer

you want the bottom to be as small as possible
that means you want the square to be as small as possible
i.e. 0
thus you want x - c^2/x =0
or x^2= c^2
x= $\pm$ c

rational · Answer

yeah, $\pm c$ is nathematically correct... but 0 or negative values for $c$ is not physically possible..

rational · Answer

https://www.desmos.com/calculator/fmm93gkum6

rational · Answer

By varying just the value of $c$, we can change the position at which the resonance occurs. This position depends only on $c$..

solomonzelman · Answer

You want the denominator to be as small as possible.
Therefore, I would try finding the minimum of.
$\color{#000000}{    \displaystyle   f(x)=\sqrt{b+\left(x-\frac{c^2}{x}\right)^2}         }$

And, based on this, I know that I am to minimize
$\color{#000000}{    \displaystyle   f(x)=b+\left(x-\frac{c^2}{x}\right)^2         }$

Because the smaller the part inside the root, the smaller the denominator, and thus the bigger the function...

And that means you are to minimize the following:
$\color{#000000}{    \displaystyle   g(x)=b+x^2-2c^2+\frac{c^4}{x^2}         }$

$\color{#000000}{    \displaystyle   g(x)=x^4+(b-2c^2)x^2+c^4         }$

You can make a substitution,
$\color{#000000}{    \displaystyle   u=x^2        }$

And use the same concept to mnimize, just as you would do by parabola (the vertex).

solomonzelman · Answer

Of course, next, you would need to sub the x back.
(This is with no calculus, with anyway seems easier than differentiating that big mess.... not that either is a problem, really.)

rational · Answer

Nice!

solomonzelman · Answer

thanks:)