Stationary Points of a graph with fractions

Question

Find the stationary points of the graph $y=\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}$ and determine the nature of each. I have differentiated but as there are x's to negative powers i dont see how it can be factorised in a way that is still usable. Any ideas?

Answer 1

HINT: use that $$f(x)=\frac{x^2+x+1}{x^3}$$ and by the quotient rule we get $$f'(x)=\frac{(2x+1)x^3-(x^2+x+1)3x^2}{x^6}$$ simplifying this we obtain $$f'(x)=\frac{-x^4-2x^3-3x^2}{x^6}$$ and we get $$f'(x)=-{\frac {{x}^{2}+2\,x+3}{{x}^{4}}}$$ this deriavtive is negative and your function monotonuoulsy decreasing for all $x\neq 0$

Answer 2

Quotient rule derivative gives

$$\dfrac{x^2+x+1}{x^3}=\dfrac{2 x +1}{3 x^2} $$

cancel $x^2$ in denominator $ x \ne 0$ , cross multiply, transpose and simplify to get

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

original expression is monotone increasing because discriminant of above quadratic equation

$$\Delta= 4-4\cdot 3\cdot 1<0 $$