Question

Prove that f(x)=x¹⁰⁰+sinx-1 is increasing on (0, 1).

Answer 1

Hello!

Let $\sf f(x) = x^{100} + sin\:x - 1$

Differentiating w.r.t x :-

$\sf f'(x)$ = $\sf 100x^{99} + cos\:x$

♦ Consider the interval ( 0, 1 ) :-

cosx > 0 and $\sf 100^{99}$ > 0 

Therefore, f'(x) > 0

Hence,
The function $\bf{f}$ is strictly increasing in interval ( 0, 1 )

Cheers!