Calculus - Showing that Min does not exists

Question

I have the set $A = \{f(x) : x > -4\}$, while $f(x)= \frac{x^2+12x+32}{x+5}$ for every $x>-4$. I want to show that $\min(A)$ does not exists.

My attempt:

I showed that: $\lim\limits_{x \to -4} f(x)=0$, so $A$ is bounded from below by $0$, and $0 \notin A$.

Now I want to show that $\inf(A)=0$, I tried using the epsilon property, and tried assuming towards contradiction, with no luck.

Any help will be amazing!

Thanks!

Answer 1

$f(x)=\frac {(x+4)(x+8)} {x+5}$ so $f(x) >0$ for all $x >-4$ and $f(x) \to 0$ as $x \to -4$. This implies that the infimum of $f$ is $0$ and this infimum is not attained.

If you want to prove that the infimum is $0$ using definition of infimum you can do the following: Since $f(x) >0$ for all $x$ it follows that $0$ is a lower bound. Let us show that no positive number is a lower bound. Let $0<\epsilon < 1$. Take $x=-4+\frac {\epsilon } 5$. Then $f(x) < \frac {(\frac {\epsilon } 5)(5)} {1}=\epsilon$ Hence, $\epsilon $ is not a lower bound. Obviously, and $\epsilon \geq 1$ also cannot be a lower bound.

Answer 2

You already know that $$\forall x>-4:\frac{x^2+12x+32}{x+5}>0$$ so that $0$ is a lower bound.

Now take $\epsilon>0$ and solve

$$\frac{x^2+12x+32}{x+5}=\epsilon$$

or, with $t:=x+4$,

$$t\frac{t+4}{t+1}=\epsilon.$$

It seems clear that, as the fraction is continuously growing from $1$, the equation always has a solution and $\epsilon$ cannot be the minimum.

Answer 3

Show that $f'(x)\gt 0 \;\;\forall x\gt -4$. It follows that $f(x)\gt 0$ for all $x\gt -4$. Hence, $0$ is a lower bound for $A$.

Claim: $0$ is $\inf A$.
Suppose not and let $t\gt 0$ be a lower bound for $A$. It follows that $t\le f(x)\;\forall x\gt -4$.
Note that $\lim_{x\to -4^+} f(x)=0$ and therefore, for every $\epsilon \gt 0, \exists \delta_\epsilon\gt 0: -4\lt x\lt -4+\delta_\epsilon \implies f(x)\lt \epsilon$.

Choose $\epsilon=t$ and get a contradiction. Therefore, $0$ is $\inf A$.

To prove that $0$ is not attained, solve $f(x)=0$ and show that there is no $x\in \mathbb R$ that satisfies the equation.

PS: $f'(x)\gt 0$ is not really required here. Note that $N^r$ and denominator of $f$ are always positive for $x\gt -4\implies f(x)\gt 0 \;\; \forall x\gt -4$.