Finding The Highest k For $k \in R$

Question

Hello everyone how can I find the highest $k \in R$ that for him

$x + k \leq 2^x$?

Answer 1

If $f(x) = 2^x-x$ then we have

$$ \lim_{x \to -\infty} f(x) = + \infty \quad \text{and}\quad \lim_{x \to +\infty} f(x) = +\infty $$ Moreover, as you said in the comments, if we differentiate we obtain $$ f'(x) = 2^x \cdot \ln 2 -1 > 0 \iff 2^x > \frac{1}{\ln 2} \iff x > -\log_2(\ln 2) $$

Hence $-\log_2(\ln 2)$ is the global minimum and

$$ \begin{split} f( -\log_2(\ln 2)) &= 2^{ -\log_2(\ln 2)}+\log_2(\ln 2)\\ &=\frac{1}{\ln 2}+\log_2(\ln 2) \\ &=\frac{1}{\ln 2}+\frac{\ln \ln 2}{\ln 2} \\ &= \frac{1 + \ln \ln 2}{\ln 2} \end{split} $$ Therefore $$k = \frac{1 + \ln \ln 2}{\ln 2}$$