Exercises: Derivatives 3 – "Math for Non-Geeks"

Mean-value theorem

Hint

From the first inequality, the transition to $-x$ for $x<1$ still allows for the inequality:

$e^{x}\leq {\frac {1}{1-x}}$

Hint

The inequality can be further extended to all $x\in \mathbb {R}$:

$|\sin(x)|\leq |x|\leq |\tan(x)|$

Where equality only holds at $x=0$.

Hint

If we even have $f(a)>g(a)$ and $f'>g'$ on $(a,b)$, then we have $f>g$ on $[a,b]$.

Hint

The generalized Bernoulli inequality can even be shown for all $x>-1$. Equality only holds in the case $x=0$.