Given, f(x) = e– ax + eax, a > 0
⇒ f'(x) = –ae–ax + aeax
⇒ f'(x) = a(eax – e– ax)
Since f(x) is monotonic increasing, so f'(x) > 0
⇒ a(eax – e–ax) > 0
⇒ a((e2ax – 1)/eax) > 0
⇒ e2ax – 1 > 0
⇒ e2ax > e0
⇒ 2ax > 0
⇒ x > 0
Hence, the required interval (0, ∞).