Taylor expansion upper bound

Question

Let $x > 0$ and let $c < 1$ be some constant. I am wondering: can I find an upper bound on $x$ such that $$ e^{-x} + (1 - c)x < 1$$ which is a function of $c$? For instance, something like $$x < 2c, \quad \text{ if } \quad c < \frac{1}{2}.$$ More generally, could people please point to relevant readings or places to learn more about this problem -- I am quite stuck.

Answer 1

You can solve analytically the equation $$e^{-x} + (1 - c)x = 1$$ Its solution is given in terms of Lambert function. Skipping the steps, the solution is $$x_*=\frac{1}{1-c}+W\left(\frac{e^{\frac{1}{c-1}}}{c-1}\right)$$ and the inequality holds for any $x < x_*$.

Close to $c=0$, using Taylor series, we have $$x_*=2 c+\frac{4 c^2}{3}+\frac{10 c^3}{9}+\frac{136 c^4}{135}+O\left(c^5\right)$$

For almost the whole range, the Padé approximant $$x_* \sim c \,\frac{2-\frac{16 }{11}c+\frac{32 }{495}c^2 } {1-\frac{46 }{33}c+\frac{67 }{165}c^2 }$$ seems to be quite good.

Answer 2

Note that $e^{-x}>0$ for any $x$. Therefore, we have $$(1-c)x<e^{-x} + (1 - c)x < 1,$$ which implies that $$x<\frac{1}{1-c}$$ since $c<1$. So we get an upper bound which is a function of $c$. But you may want some other bound.