# Differentiation; Maxima and Minima

Suppose $y=\frac{1}{x(1+n^{2}x^{2})}$ . Then when will y be maximum and what its maximum value. I don't know how to differentiate it so that i can find the maximum value. Please help me.

• To differentiate, use the fact that $1/f(x) = [f(x)]^{-1}$ (not to be confused with a functions inverse though) and then use the power and product rules. – Mattos Oct 12 '16 at 14:37
• Note that $y$ is undefined when $x=0$. Since $\lim_{x\to0+}y(x)=\infty$ and $\lim_{x\to0-}y(x)=-\infty$ the variable (function) $y$ assumes neither a minimum nor a maximum on its domain of definition. – Christian Blatter Oct 12 '16 at 15:33

I assume the variable is $x$, and $n$ is any real constant (an integer will do, but the following is valid for any real $n$).

If $x$ is free to take any value in $]-\infty,0[\,\cup\,]0,+\infty[$ (as $y$ is not defined for $x=0$), then there is no minimum value. Just pick an arbitrary small positive number $\epsilon$, and let $x=-\epsilon$, then $y$ takes on arbitrarily large negative values. There is no infimum in the reals, nor a minimum, since the set of values of $y$ has no lower bound.

If $x$ is restricted to positive values, that is $x>0$, then there is no minimum either, but there is an infimum: let $x$ be an arbitrarily large positive number, then $y$ takes on arbitrarily small positive values, but never reaches $0$, so the inf is zero, but it is not reached so the minimum does not exist.

Last, there is no maximum. Let $x=\epsilon$, then $y$ takes on arbirarily large positive values. There is no supremum in the reals, nor a maximum, since the set of values of $y$ has no upper bound.

HINT

If u,v are functions of x and when v/v is a maximum or constant, applying quotient rule

$$\frac{u}{v}= \frac{v du - u dv} {v^2}=0$$

gives

$$\frac{u}{v}= \frac{du/dx} {dv/dx}$$

or

$$\frac{1}{x(1+ n^2 x^2)}= \frac{0} {1+ 3 n^2 x^2}$$

on simplification

$$1+ 3 n^2 x^2 =0$$

which can have no real roots.No max/min for a monotonic function.