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For a bounded real function on $[a,b]$, take a partition $a=t_0<\cdots <t_m=b$. Its lower and upper Darboux sums areThe function is Riemann integrable when ${\sup }_{P}L(f,P)={\inf }_{P}U(f,P)$; the common value is its Riemann integral. Equivalently, for every $\epsilon >0$ there is a partition with $U(f,P)-L(f,P)<\epsilon$. A continuous function is bounded and Riemann integrable on each compact interval, so all the following integrals exist.

Set $h(x)=x^{-1}{\int }_0^{x}f(t)dt$ for $x>0$. Strict decrease gives $f(x)<f(t)<f(0)$ for $0<t<x$. Integrating, with strict inequalities because the difference is positive on interior subintervals, givesThe intermediate value theorem produces $g(x)\in (0,x)$ with $f(g(x))=h(x)$, and strict monotonicity makes this point unique. Thus there is exactly one such $g(x)$ in $(0,x)$.

For the given exponential function, $h(x)=(1-e^{-x})/x$, soThe preceding strict bounds already prove $0<g(x)<x$; no limiting value at zero needs to be included in its domain.

For differentiability, a continuous strictly monotone function has a continuous inverse on its range. If it is differentiable at an interior point $t$ with $f^{\prime }(t)\ne 0$, the inverse is differentiable at $y=f(t)$, with derivative $1/f^{\prime }(t)$. Indeed, setting $u=f^{-1}(y^{\prime })$, inverse continuity gives $u\to t$ as $y^{\prime }\to y$, andThis one-dimensional inverse result needs no assumption that $f^{\prime }$ is continuous. It is enough here that $f$ is differentiable and $f^{\prime }<0$ at every positive argument.

The fundamental theorem of calculus and the quotient rule giveNow $g=f^{-1}\circ h$ and $g(x)>0$, so the inverse derivative formula applies at every relevant point. The monotone mean-value point of an integral average satisfiesBoth numerator and denominator are negative. For the exponential example, this also agrees with $g^{\prime }(x)=1/x-1/(e^x-1)>0$, since $e^x>1+x$ for $x>0$.