The function $f ( x ) = ( x + 1 ) ^ { 2 }$ is one-to-one if its domain is

A) $( - \infty , - 1 ]$
B) $( - \infty , \infty )$
C) $[ - 2 , \infty )$
D) $( - \infty , - 1 ) \cup ( - 1 , \infty )$
E) $( - \infty , 1 ]$

Question

The function $f ( x ) = ( x + 1 ) ^ { 2 }$ is one-to-one if its domain is

A) $( - \infty , - 1 ]$
B) $( - \infty , \infty )$
C) $[ - 2 , \infty )$
D) $( - \infty , - 1 ) \cup ( - 1 , \infty )$
E) $( - \infty , 1 ]$

Quizplus · Accepted Answer

The function $f(x) = (x + 1)^2$ is not one-to-one over its entire domain because it is a quadratic function, which is symmetric about its vertex. The vertex of this function is at $(-1, 0)$. Restricting the domain to $(- \infty, -1]$ makes the function one-to-one because in this interval, the function is either strictly increasing or constant (in this case, it's constant at the single point $x = -1$). This restriction eliminates the symmetry issue, ensuring that each value of $f(x)$ corresponds to exactly one value of $x$ in the given domain.

The Function f(x)=(x+1)2f ( x ) = ( x + 1 ) ^ { 2 }f(x)=(x+1)2

The Function $f ( x ) = ( x + 1 ) ^ { 2 }$