Question

$f(x)=e^{\{x\}}$

number of Integers in Range of $f(x)=?$

Answer 1

2

Answer 2

The function is given by $f(x) = e^{\{x\}}$, where $\{x\}$ denotes the fractional part of $x$.

The range of the fractional part function $\{x\}$ is $[0, 1)$. This means for any real number $x$, $0 \le \{x\} < 1$.

The base of the exponential function is $e$, which is approximately $2.718$. Since $e > 1$, the exponential function $g(y) = e^y$ is strictly increasing for all real values of $y$.

To find the range of $f(x)$, we evaluate the function $e^y$ over the range of the exponent $\{x\}$, which is $[0, 1)$.

As $\{x\}$ varies from $0$ up to (but not including) $1$, $e^{\{x\}}$ varies from $e^0$ up to (but not including) $e^1$.

$e^0 = 1$ and $e^1 = e$.

Therefore, the range of $f(x)$ is $[1, e)$.

We are asked to find the number of integers in this range $[1, e)$.

We need to find the integers $n$ such that $1 \le n < e$.

Since $e \approx 2.718$, the inequality becomes $1 \le n < 2.718$.

The integers that satisfy this inequality are $n=1$ and $n=2$.

The next integer is $3$, but $3 > e$, so it is not in the range.

The integers in the range $[1, e)$ are 1 and 2.

The number of such integers is 2.