Let's break down the problem. The function provided is:

$f(x) = 2x + 1, \text{ for } x < 0$

The function is only defined for $x < 0$, meaning we can only evaluate $f(x)$ for values less than zero.

Now, let's evaluate each part:

(a) $f(-1)$

We need to find $f(-1)$. Since $-1 < 0$, we can apply the function rule:

$f(-1) = 2(-1) + 1 = -2 + 1 = -1$

So, $f(-1) = -1$.

(b) $f(0)$

The function is only defined for $x < 0$. Since $0 \not< 0$, $f(0)$ is undefined.

(c) $f(2)$

Similarly, $2 \not< 0$, so $f(2)$ is undefined as well.

(d) $f(t^2 + 1)$

Here, we are asked to evaluate $f(t^2 + 1)$. Since $t^2 + 1$ is always positive (for all real values of $t$), and the function is only defined for $x < 0$, $f(t^2 + 1)$ is also undefined.

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Domain

The domain of $f(x)$ is the set of $x$-values for which the function is defined. In this case, the function is only defined for $x < 0$, so the domain is:

$\text{Domain: } (-\infty, 0)$

Range

To find the range, we analyze the output values of $f(x) = 2x + 1$ for $x < 0$. As $x$ approaches $0$ from the left (i.e., $x \to 0^-$), $f(x)$ approaches $1$. As $x$ becomes more negative (i.e., $x \to -\infty$), $f(x)$ decreases without bound. Thus, the range is:

$\text{Range: } (-\infty, 1)$

Summary of Results:

• $f(-1) = -1$
• $f(0)$ is undefined
• $f(2)$ is undefined
• $f(t^2 + 1)$ is undefined

Domain: $(- \infty, 0)$

Range: $(- \infty, 1)$

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Do you need more details or have any questions?

Here are 5 related questions:

1. How would the domain change if the function were defined for $x \geq 0$?
2. What happens to the range if we modify the function to $f(x) = 2x - 1$?
3. How can we check if a function is continuous at a specific point?
4. What is the interpretation of a function being undefined for certain values of $x$?
5. How does the slope of the function $f(x) = 2x + 1$ affect its graph?

Tip: The domain of a function is the set of input values that don't violate any conditions set by the function's definition (e.g., restrictions on $x$ like $x < 0$ or square roots requiring non-negative arguments).