19. if f(x) = (e^x / x)+1, determine the interval(s) on which f is decreasing.

Explanation:

Step1: Find the derivative of $f(x)$

Use the quotient - rule. If $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = e^{x}$, $u^\prime=e^{x}$, $v = x$, $v^\prime = 1$, and $f(x)=\frac{e^{x}}{x}+1$. So $f^\prime(x)=\frac{e^{x}\cdot x - e^{x}\cdot1}{x^{2}}=\frac{e^{x}(x - 1)}{x^{2}}$.

Step2: Find where $f^\prime(x)<0$

Set $f^\prime(x)<0$, i.e., $\frac{e^{x}(x - 1)}{x^{2}}<0$. Since $e^{x}>0$ for all real $x$ and $x^{2}>0$ for all $x\neq0$, we only need to consider the sign of $x - 1$. Solving $x - 1<0$ gives $x<1$. But we must exclude $x = 0$ (because the original function $f(x)=\frac{e^{x}}{x}+1$ is undefined at $x = 0$).

Answer:

$(-\infty,0)\cup(0,1)$