Calculus easy 1

Just look that they are the inverse of each other and monotonic too so they have the solution only at $f(x) = x$ The rest is an easy part.

${e}^{x-k} = x$

Taking natural log on each side

$x-k = \ln{x}$

$k = x-\ln{x}$

$k = \ln{\dfrac{{e}^{x}} {x}}$ at $x=1$ {minimum holds check by differentiation} $\dfrac {{e}^{x}} {x} \ge e$

Since $\ln{x}$ is an increasing function,

$\ln{\dfrac {{e}^{x}} {x}} \ge 1$

$k \ge 1$

Again, ${e}^{x}$ is an increasing function,

So ${e}^{k} \ge {e}^{1}$