To determine whether the function \( f(x) = x^2 + 2x + 1 \) is one-to-one, we need to check if, for every pair of distinct inputs \( a \) and \( b \) in the domain of the function, \( f(a) \) is not equal to \( f(b) \).
A function is one-to-one if different inputs lead to different outputs. This can be confirmed using either the definition of a one-to-one function, or we can analyze the function using calculus or by inspecting its algebraic properties.
First, let's rewrite the function:
\[ f(x) = x^2 + 2x + 1 = (x + 1)^2 \]
This shows that the function is a perfect square. The graph of a perfect square function is a parabola that opens upwards and has its vertex as the minimum point.
The vertex occurs at the point where \( x = -b/(2a) \). For our function \( x^2 + 2x + 1 = 1(x^2) + 2(x) + 1 \), we identify \( a = 1 \) and \( b = 2 \):
\[ x = -\frac{2}{2 \cdot 1} = -1 \]
The vertex is at \( (-1, 0) \), which is the minimum point of the parabola. For every value of \( x \) less than -1, the function will produce increasing values, and for every value greater than -1, the function will also produce increasing values. This results in the same output for two different inputs on either side of the vertex:
- For example, \( f(-2) = (-2 + 1)^2 = (-1)^2 = 1 \)
- And \( f(0) = (0 + 1)^2 = 1^2 = 1 \)
Since \( f(-2) = f(0) \) but \(-2 \neq 0\), we can conclude that this function is not one-to-one.
To summarize, the function \( f(x) = x^2 + 2x + 1 \) is not one-to-one because it produces the same output for different inputs.