Let $f:\mathbb R\to \mathbb R$ be differentiable such that $f'(x)>f(x)$ for all $x\in \mathbb R$ and $f(0)=1$, then $f(1)$ lies in which interval?

Question

Let $f:\mathbb R\to \mathbb R$ be differentiable function such that $f'(x)>f(x)$ for all $x\in \mathbb R$ and $f(0)=1$, then $f(1)$ lies in which one of the intervals ?

a)$(0,e^{-1})$

b)$(e^{-1},\sqrt{e})$

c) $(\sqrt{e},e)$

d)$(e,\infty)$

This is how I tried this question:

let $f(x)= 2e^{x}-1, $ so $f(0)=1$

$f'(x)=2e^{x} >f(x) \forall \; x\in R$

so $f$ satisfies the given conditions,

$f(1)=2e-1>e$ so d) should be correct. Is this correct? How can I solve this without using a particular example?

Answer 1

Let $g(x)=e^{-x}f(x)$. Then $g'(x)=e^{-x} (f'(x)-f(x)) >0$. This makes $g$ increasing . Hence $g(1) >g(0)$ which gives $f(1) >e$. Hence d) is true. This automatically rules out the other options.