Find the range for $ f(x) = \cfrac{e^x}{1+\lceil x \rceil } $ when $x\ge 0$
My book answers it in a very straight forward manner -
Here $f(x)$ is defined for all $x \ge 0$ . Also, $f(x)$ is an increasing function in $[0,\infty)$. Thus, range = $[f(0),f(\infty)] = [1, \infty)$
My question is :
How is this function an increasing function in $[0, \infty)$ ?
You have a very good question. For instance, for $x$ just below $1$, we have that $$ f(0.999) = \frac{e^{0.999}}{1} \approx e,$$ while for $x = 1$ we have $$ f(1) = \frac{e}{2} < e.$$ So the function is not increasing. But it's not so hard to see that it will always be at least $1$, and it becomes unboundedly large. And at integer values, it temporarily decreases, so that every value in $[1, \infty)$ gets hit. I think this is the way to prove the range is $[1, \infty)$.