$y = xe^{−1/|x|}$ for $−∞ < x < ∞$

Question

(a) Let $f(x) = x − xe^{−1/x}, x > 0$. Show that f(x) is an increasing function on $(0,∞),$ and $\lim_{x→∞} f(x) = 1$.

(b) Using part (a) and calculus, sketch the graphs of $y = x−1, y = x, y = x + 1,$ and $y = xe^{−1/|x|}$ for $−∞ < x < ∞$ using the same $X $and $Y$ axes.

My approach: for (a) I could only show $\lim_{x→∞} f(x) = 1$. I proved it using the convergent series expansion of $e^x$.

for (b) the problem lies with $y = xe^{−1/|x|}$. I am not being able to sketch it.

$f'(x)=1-{e^{-1\over x}\over x}-e^{-1\over x}$

Answer 1

For (a), proving that the function is increasing is usually easiest if you prove that its derivative is positive.

For (b), the usual things you must show in a graph are:

• asymptotes (behaviours of the function as $x$ goes to $\infty$ or $-\infty$).
• minimums and maximums
• zeroes (if there are any).

For each of these things, some calculation is required.

Answer 2

You already have the derivative: $$ f^{\prime }(x)=1-\frac{\exp [-\frac{1}{x}]}{x}-\exp [-\frac{1}{x}]. $$ Now write it as $$ f^{\prime }(x)=1-\frac{1+\frac{1}{x}}{\exp [\frac{1}{x}]}. $$ Since $$ \exp [\frac{1}{x}]>1+\frac{1}{x},\;0<x<\infty $$ $f^{\prime }(x)$ is positive for such $x$. For $x=0$, $f^{\prime }(x)=1$, whereas for $x\rightarrow \infty, $ $f^{\prime }(x)\rightarrow 0$.