A calculus problem by Danish Ahmed

Idea from @Aakash Khandelwal , presented more clearly.

For $x \ge 1$ , the nominator and denominator of the expression are non-negative, therefore, if the largest value of $\dfrac {x^4-x^2}{x^6+2x^3-1} = \dfrac 1a$ , then the smallest value of its reciprocal $\dfrac {x^6+2x^3-1}{x^4-x^2} = a$

$\begin{aligned} f(x) & = \frac {x^6+2x^3-1}{x^4-x^2} & \small \color{#3D99F6}{\text{Divide up and down by }x^3} \\ & = \frac {x^3+2 - \frac 1{x^3}}{x-\frac 1x} \\ & = \frac {\left(x - \frac 1x\right) \left(x^2+1+\frac 1{x^2}\right)+2}{x-\frac 1x} \\ & = \frac {\left(x - \frac 1x\right) \left(x^2-2+\frac 1{x^2}\right)+3\left(x - \frac 1x\right)+2}{x-\frac 1x} \\ & = \frac {\left(x - \frac 1x\right)^3+3\left(x - \frac 1x\right)+2}{x-\frac 1x} \\ & = \left(\color{#3D99F6}{x - \frac 1x} \right)^2 + 3 + \frac 2{\color{#3D99F6}{x - \frac 1x}} \end{aligned}$

Let $\color{#3D99F6}{u = x - \dfrac 1x}$ , we have:

$\begin{aligned} f(\color{#3D99F6}{u}) & = \color{#3D99F6}{u}^2 + 3 + \frac 2{\color{#3D99F6}{u}} \\ f'(u) & = 2u - \frac 2{u^2} & \small \color{#3D99F6}{\text{Put }f'(u) = 0} \\ \implies u & = 1 \end{aligned}$

$\implies a = f_{min}(x) = f_{min}(u) = 1^2 + \dfrac 51 = \boxed{6}$

Take out $x^{3}$ common from both numerator and denominator .

Put $x-1/x$ = $t$ .

Now the given expression is

$t/(t^{3}+3t+2$

Differentiate it to get critical point as $t=1$

Answer follows after that.

To find the maximum of a function, we can differentiate the function and solve when the derivative is equal to 0. Differentiation is as follows:

$f(x)=\frac {x^4-x^2}{x^6+2x^3-1}$

$f'(x)=\frac {d}{dx} \frac {x^4-x^2}{x^6+2x^3-1}$ $f'(x)=-2x^9 + 4x^7 + 2x^6 + 2x^4 - 4x^3 + 2x$

Solving for when f'(x) = 0, we arrive at $x=1.618$ . We can now be sure this step was correct, because 1.618 is $>= 1$ , as specified in the problem. Substitute this value into the original function:

$\frac {x^4-x^2}{x^6+2x^3-1}$

$\frac {(1.618)^4-(1.618)^2}{(1.618)^6+2(1.618)^3-1} = \frac 16$

Since this is equal to $\frac {1}{a}$ , $a = 6$ .