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. $f(-4) = \sqrt{17}$
(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. $f(-10) = [ ]$
B. The square root is not a real number
1. **State the problem:** We need to find the values of the function $f(t) = \sqrt{t^2 + 1}$ for $t=0$, $t=-4$, and $t=-10$.
2. **Recall the formula:** The function is defined as $f(t) = \sqrt{t^2 + 1}$. Since $t^2$ is always non-negative and $1$ is positive, the expression inside the square root is always positive or zero, so the square root will always be a real number.
3. **Evaluate $f(0)$:**
$$f(0) = \sqrt{0^2 + 1} = \sqrt{0 + 1} = \sqrt{1} = 1$$
4. **Evaluate $f(-4)$:**
$$f(-4) = \sqrt{(-4)^2 + 1} = \sqrt{16 + 1} = \sqrt{17}$$
5. **Evaluate $f(-10)$:**
$$f(-10) = \sqrt{(-10)^2 + 1} = \sqrt{100 + 1} = \sqrt{101}$$
6. **Conclusion:** All values exist and are real numbers.
**Final answers:**
- $f(0) = 1$
- $f(-4) = \sqrt{17}$
- $f(-10) = \sqrt{101}$