Functions Question 371
Question: The range of the function $ f(x)=x^{2}+2x+2 $ is
Options:
A) $ (1,\infty ) $
B) $ (2,\infty ) $
C) $ (0,\infty ) $
D) $ [1,\infty ) $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let $ f(x)=y=x^{2}+2x+2={{(x+1)}^{2}}+1 $ $ y-1={{(x+1)}^{2}}\Rightarrow x+1=\sqrt{y-1} $ $ x=\sqrt{y-1}-1 $ Since, $ y-1\ge 0\therefore y\ge 1 $
$ \therefore $ range is $ [ 1,\infty ) $ .
BETA
