Question: Find each function value, if it exists.
f(t) = \sqrt{t^2 + 1}
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. f(0) = 1 (Simplify your answer. Type an exact answer, using radicals as needed.)
B. The square root is not a real number.
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. f(-4) = \square
(Simplify your answer. Type an exact answer, using radicals as needed.)
B. The square root is not a real number.
1. **State the problem:** We are given the function $$f(t) = \sqrt{t^2 + 1}$$ and need to find the values of $$f(0)$$ and $$f(-4)$$ if they exist.
2. **Recall the domain and formula:** The expression inside the square root, called the radicand, must be non-negative for the function to have real values. Since $$t^2 \geq 0$$ for all real $$t$$, and $$1 > 0$$, the radicand $$t^2 + 1 \geq 1 > 0$$ for all real $$t$$. So the function is defined for all real numbers.
3. **Calculate $$f(0)$$:**
$$f(0) = \sqrt{0^2 + 1} = \sqrt{0 + 1} = \sqrt{1} = 1$$
4. **Calculate $$f(-4)$$:**
$$f(-4) = \sqrt{(-4)^2 + 1} = \sqrt{16 + 1} = \sqrt{17}$$
Since $$\sqrt{17}$$ is a real number, the function value exists.
**Final answers:**
- $$f(0) = 1$$
- $$f(-4) = \sqrt{17}$$