Question

Find the minimum value of \(\frac{(x+1 / x)^{6}-\left(x^{6}+1 / x^{6}\right)-2}{(x+1 / x)^{3}+\left(x^{3}+1 / x^{3}\right)}\) for \(x>0\)

Short answer

The minimum value of given expression for \(x > 0\) is -0.25.

Step-by-step solution

Simplify The Expression

Rewrite every term in the expression in terms of \(x+1/x\) and \(x^3+1/x^3\). The resulting expression reads as: \(\frac{(x+1/x)^6-(x+1/x)(x^3+1/x^3)(x^3+1/x^3)-2}{(x+1/x)^3+(x^3+1/x^3)}\) which simplifies to: \(\frac{(x+1/x)^3-(x^3+1/x^3)-2}{(x+1/x)^3+(x^3+1/x^3)}\)

Further Simplification

Simplifying even further gives: \(\frac{(x+1/x)^3-(x^3+1/x^3)-2}{2(x+1/x)(x^3+1/x^3)}\)

Substitute Values

Substitute \(y=x+1/x\) and \(z=x^3+1/x^3\), then the expression becomes: \(\frac{y^3 - z - 2}{2yz}\)

Derive the New Equation

The derivative of this new equation, by using the quotient rule for derivatives \[f'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{[h(x)]^2}\] becomes: \[dz=\frac{6y(y^2-z-1)}{y^2}\] when \(y\ne0\)

Find the Critical Points

Identify the critical points where dz is equal to zero: \[y^2-z-1=0\] \[y^2-(y^2-1)-1=0\] \[y^2-y^2=0\] \[0=0\] Therefore y has no real roots.

Obtain Minimum Value of the Expression

The minimum value of the expression is attained when \(y = -b/2a\). Here \(a = 2\), and \(b = 1\), hence \(y = -1/4 = -0.25\) is a minimum value.