Question

Let $f(x)=x^{2}+2x+3,x\le-1$. Then the domain of the inverse of $f(x)$ is

• A. $[-2,\infty)$
• B. $[12,\infty)$
• C. $[-22,\infty)$
• D. $[-12,\infty)$
• E. $[2,\infty)$

Hint:

Math Tip: Always remember the golden rule of inverse functions: the $x$ and $y$ values swap.
Domain of $f^{-1} = $ Range of $f$
Range of $f^{-1} = $ Domain of $f$
Completing the square is the fastest way to find the range of a quadratic function.

Answer 1

Answer: (E)

Concept:
Functions - Domain and Range of Inverse Functions.
Step 1: Recall the relationship between a function and its inverse.
The domain of an inverse function $f^{-1}(x)$ is exactly equal to the range of the original function $f(x)$.
Therefore, our objective is to find the range of $f(x)$ for the given domain $x \le -1$.
Step 2: Rewrite the function by completing the square.
Given $f(x) = x^2 + 2x + 3$.
Rewrite it as $f(x) = (x^2 + 2x + 1) + 2$.
$$ f(x) = (x + 1)^2 + 2 $$ This is the vertex form of the parabola, showing a minimum point at $(-1, 2)$.
Step 3: Determine the behavior of the function on the given domain.
The given domain restricts $x$ to $x \le -1$.
On this interval, the squared term $(x + 1)^2$ will always be $\ge 0$.
At exactly $x = -1$, the function reaches its minimum value: $$ f(-1) = (-1 + 1)^2 + 2 = 0 + 2 = 2 $$ As $x$ takes values less than $-1$ (e.g., $-2, -3, -\infty$), the value of $(x+1)^2$ increases towards infinity.
Step 4: Determine the range of f(x) and the domain of its inverse.
Since the minimum value is $2$ and it goes up to infinity, the range of $f(x)$ is $[2, \infty)$.
Because Domain$(f^{-1}) = $ Range$(f)$, the domain of the inverse function is $[2, \infty)$.