Question

Prove x¹⁰¹+sinx-1 is increasing for | x | > 1.

Answer 1

we have to prove that f(x) = x¹⁰¹+sinx-1, is increasing for |x| > 1

first differentiate f(x) with respect to x,

f'(x) = 101x¹⁰⁰ + cosx

we know, x¹⁰⁰ always will be positive for all |x| > 1

i.e., x¹⁰⁰ = |x|¹⁰⁰ > 1

⇒101x¹⁰⁰ > 101

and we also know, value of cosine function always lies between -1 to 1.

so, -1 ≤ cosx ≤ 1

so, 101x¹⁰⁰ + cosx > 0

hence, f'(x) = 101x¹⁰⁰ + cosx > 0

therefore, f(x) = x¹⁰¹ + sinx - 1, is increasing for |x| > 1