Given the function $f(x) = (x-1)^2 - 4$ with the domain $f : [2, \infty) \to [-4, \infty)$, let's address each part of the question:

1. Show that the function is well-defined

To show that the function is well-defined, we must check that for every $x$ in the domain $[2, \infty)$, the function produces an output in the range $[-4, \infty)$.

The function is $f(x) = (x-1)^2 - 4$.

• Step 1: Calculate the value of $f(x)$ at the boundary of the domain, $x = 2$: $f(2) = (2-1)^2 - 4 = 1^2 - 4 = 1 - 4 = -3$ Therefore, $f(2) = -3$.

• Step 2: For $x$ approaching $\infty$, $(x-1)^2$ becomes large, so: $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} (x-1)^2 - 4 = \infty - 4 = \infty$ This shows the function's output can reach up to $\infty$.

Thus, $f(x)$ maps $[2, \infty)$ to $[-4, \infty)$, making the function well-defined.

2. Sketch the graph of the function

The graph of the function $f(x) = (x-1)^2 - 4$ is a parabola opening upwards with its vertex at $(1, -4)$. Since the domain starts at $x = 2$, the graph should start from the point $(2, -3)$ and move upwards to infinity.

3. Find the image of the function

The image (range) of the function is the set of values $f(x)$ can take as $x$ varies over its domain $[2, \infty)$. From the analysis above:

• The minimum value of $f(x)$ occurs at $x = 2$, which is $-3$.
• As $x \to \infty$, $f(x) \to \infty$.

Thus, the image of the function is $[-3, \infty)$.

4. Determine if the function is injective (one-to-one)

To check if the function is injective, we need to verify if different inputs map to different outputs.

The derivative of the function $f(x) = (x-1)^2 - 4$ is: $f'(x) = 2(x-1)$ For $x \geq 2$, $x-1 \geq 1$, so $f'(x) \geq 2$. Since the derivative is positive for all $x \geq 2$, the function is strictly increasing and thus injective.

5. Determine if the function is surjective (onto)

To determine if the function is surjective, we need to check if every element in the codomain $[-4, \infty)$ has a corresponding preimage in $[2, \infty)$.

Given the calculated image $[-3, \infty)$, the function is not surjective because the interval $[-4, -3)$ in the codomain has no preimage in the domain.

6. Prove the function is monotonically increasing

We already calculated the derivative $f'(x) = 2(x-1)$. Since $f'(x) > 0$ for $x \geq 2$, $f(x)$ is monotonically increasing on $[2, \infty)$.

This completes the analysis of the function $f(x)$.

Would you like a more detailed sketch of the graph or any further clarification on these points?

Related Questions

1. How does the function behave as $x$ approaches the lower boundary of the domain?
2. What is the significance of the vertex in the context of this function's behavior?
3. How can we formally verify the injectivity of a different function?
4. What methods can be used to find the inverse of a function like this?
5. Can you provide an example where the function is both injective and surjective?
6. How does the shape of the parabola change if we modify the function to $f(x) = a(x-1)^2 - 4$ with $a > 0$?
7. What role does the range $[-4, \infty)$ play in defining the function?
8. How does the derivative help in understanding the function's monotonicity?

Tip:

When analyzing the monotonicity of a function, always check the sign of the derivative across the domain. This will help determine whether the function is strictly increasing, strictly decreasing, or neither.