I'm required to compare $e^{4}-2$ and $50$ without using calculator. I thought of the following way:
Let a function $h(x)$ be defined as $e^{x}-13x.$ If I can prove that this function at $x=4$ is positive(without the use of calculator again) then the job is done. But I'm not able to do so. How to proceed?
Best Answer
Use:
$$e^2>1+\frac21+\frac{2^2}2+\frac{2^3}6+\frac{2^4}{24}+\frac{2^5}{120}=7 +\frac{4}{15}.$$ So $$e^4>\left(7+\frac4{15}\right)^2>7^2+2\cdot 7\cdot \frac{4}{15}>52.$$
The last step because: $$2\cdot 7\cdot\frac4{15} =(15-1)\cdot \frac4{15}=4-\frac{4}{15}>3.$$