Find the minimum and maximum value of $h(x)$

Question

Let $$f(x)= x^2 +\frac1{x^2}$$ and $$g(x)=x-\frac1x.$$

Then how do I myself find the maximum and minimum value of $h(x)=f(x)/g(x)$?

Answer 1

Well first notice that $g(x)$ has a zero at $x=1$, and at this point $f(x)$ is finite and non-zero. $f(x)$ is always positive. If we approach $x=1$ from the right $g(x)$ is positive, and if we approach $x=1$ from the left $g(x)$ is negative. Hence depending from which direction we approach $x=1$, $h(x)$ goes to $+\infty$ and $-\infty$, making the function unbounded.

Answer 2

You may just do the calculations to get

$$h(x) = \frac{f(x)}{g(x)} = \frac{x^2 +1/x^2}{x-1/x}=\frac{x^4+1}{x(x^2-1)}$$

Now, you can see that when

$x$ tends to $1^+$, $-1^-$, or $0^-$, $\hspace{1cm} h(x)$ tends to $+\infty$,

and when

$x$ tends to $1^-$, $-1^+$, or $0^+$, $\hspace{1cm} $ $h(x)$ tends to $-\infty$.

Answer 3

Refer to the graph:

$\hspace{0.5cm}$

Note: $$\lim_{x\to-\infty} h(x)=\lim_{x\to-1^-} h(x)=\lim_{x\to0^+} h(x)=\lim_{x\to1^-} h(x)=-\infty;\\ \lim_{x\to-1^+} h(x)=\lim_{x\to0^-} h(x)=\lim_{x\to1^+} h(x)=\lim_{x\to+\infty} h(x)=+\infty.$$

Answer 4

Note that $$h(x) = \frac{x^2 +\frac1{x^2}}{x-\frac1x}=\frac{x+\frac1{x^3}}{1-\frac1{x^2}}\to\pm\infty$$ as $x\to\pm\infty$.