Functions Part 1

Given that f(x)=(x+1)2−1,x≥−1
Clearly Df=[−1,∞) but co-domain is not given
∴f(x) need not be necessarily onto.
But if f(x) is onto then as f(x) is one one also,
(x+1) being something +ve,f−1(x) will exist where
‌(x+1)2−1=y
‌⇒x+1=√y+1
(+ve square root as x+1≥0)
‌⇒x=−1+√y+1
‌⇒f−1(x)=√x+1−1
Then ‌‌f(x)=f−1(x)
‌⇒(x+1)2−1=√x+1−1
‌⇒(x+1)2=√x+1
‌⇒(x+1)4=(x+1)
‌⇒(x+1)[(x+1)3−1]=0
‌⇒x=−1,0
∴ The statement −1 is correct but statement- 2 is false.