Best way to explain how the Infimum and Supremum of this function are obtained...?

Question

I have the function $\;f(x)=\dfrac{x^{(1/2)}}{2+x}\;$ and I know that $\inf(f)$ does not exist and $\sup(f)=2$ but I don't know how to formally show this rigorously? Anyone got a formal way of showing this, it would be much appreciated...

Answer 1

$$f(x) = \dfrac{x^{1/2}}{2 + x}$$

• Domain of $f(x)$: $[0, +\infty) \subset \mathbb R$.

• We find critical points (and associated extrema) by finding $f'(x)$
  and solving where $f'(x) = 0$.

  $$f'(x) = \dfrac{\frac 12x^{-1/2}(2+x) - x^{1/2}}{(2 + x)^2} =\dfrac{ (2+x) - 2x}{2x^{1/2}(2 + x)^2} = \dfrac{2 - x}{2\sqrt x(x+ 2)^2}$$

  $$f'(x) = 0 \iff x = 2;\quad x = 2 \implies f(x) = \dfrac{\sqrt 2}{4} = \dfrac 1{2 \sqrt 2}$$

  • Confirm, using the derivative, the point $(2, \frac 1{2\sqrt 2})$ is a global maximum of $f(x)$.

• Note also that $$\lim_{x\to 0^+} \dfrac{x^{1/2}}{2 + x} = 0/2 = 0$$ and $$\lim_{x\to +\infty} \dfrac{x^{1/2}}{2 + x} = \lim_{x\to +\infty}\dfrac{\frac{1}{x^{1/2}}}{\frac 2x + 1} = 0$$

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$$\therefore \underbrace{0}_{\inf} \leq f(x) \leq \underbrace{\dfrac 1{2\sqrt 2}}_{\sup}$$