Problem 1. Let $f$ and $g$ denote functions defined on some set $A$.

Prove that
$$\sup {x \in A}(f(x)+g(x)) \leq \sup {x \in A} f(x)+\sup _{x \in A} g(x) .$$

Find an example for a pair $f, g$ for which
$$\sup {x \in A}(f(x)+g(x))=\sup {x \in A} f(x)+\sup _{x \in A} g(x) .$$

Find an example for a pair $f, g$ for which
$$\sup {x \in A}(f(x)+g(x))<\sup {x \in A} f(x)+\sup _{x \in A} g(x) .$$

Solution.

For any $x \in A, f(x) \leq \sup {x \in A} f(x)$ and $g(x) \leq \sup {x \in A} g(x)$ and hence $f(x)+g(x) \leq$ $\sup {x \in A} f(x)+\sup {x \in A} g(x)$. Thus $\sup {x \in A} f(x)+\sup {x \in A} g(x)$ is an upper bound for $f(x)+g(x)$ on $A$, and hence it is no smaller than the least upper bound for $f(x)+g(x)$ on $A$, which is $\sup _{x \in A}(f(x)+g(x))$.

Take say $f$ and $g$ to be the constant functions 0 , and then $\sup {x \in A}(f(x)+g(x))$ and $\sup {x \in A} f(x)+\sup _{x \in A} g(x)$ are both 0.

Take say $f(x)=x$ and $g(x)=-x$ on $A=[0,1]$. Then $f(x)+g(x)=0$ and hence $\sup {x \in A}(f(x)+g(x))=0$ while $\sup {x \in A} f(x)=1$ and $\sup {x \in A} g(x)=0$ and hence $\sup {x \in A} f(x)+\sup {x \in A} g(x)=1$. Thus $\sup {x \in A}(f(x)+g(x))=0<1=\sup {x \in A} f(x)+$ $\sup {x \in A} g(x)$ as required.

Problem 2. Sketch the graph of the function $y=f(x)=\frac{x^2}{x^2-1}$. Make sure that your graph clearly indicates the following:

The domain of definition of $f(x)$.

The behaviour of $f(x)$ near the points where it is not defined (if any) and as $x \rightarrow \pm \infty$.

The exact coordinates of the $x$ – and $y$-intercepts and all minimas and maximas of $f(x)$.

Solution. $f(x)$ is defined for $x \neq \pm 1$, and the following limits are easily computed: $\lim {x \rightarrow \pm \infty} f(x)=1, \lim {x \rightarrow-1^{-}} f(x)=\lim {x \rightarrow 1^{+}} f(x)=\infty$ and $\lim {x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=$ $-\infty$. The only solution for $f(x)=0$ is $x=0$, hence the only intersection of the graph of $f(x)$ with the axes is at $(0,0)$. Other than at $x=0$, the numerator of $f$ is always positive, hence the sign of the function is determined by the sign of the denominator $x^2-1$. Thus $f(x) \leq 0$ for $|x|<1$ and $f(x)>0$ for $|x|>1$. Finally $f^{\prime}(x)=\frac{2 x\left(x^2-1\right)-x^2 2 x}{\left(x^2-1\right)^2}=-\frac{2 x}{\left(x^2-1\right)^2}$ and thus $f^{\prime}$ is positive and $f$ is increasing (locally) for $x<0$ and $f^{\prime}$ is negative and $f$ is decreasing (locally) for $x>0$. Thus overall the graph is:

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