Find the domain of the function.
$f ( t ) = \sqrt { 64 - t ^ { 2 } }$
A) $- 8 \leq t \leq 8$
B) $t \leq - 8$ or $t \geq 8$
C) $t \geq 0$
D) $t \leq 8$
E) all real numbers

Question

Quizplus · Accepted Answer

The domain of $f(t) = \sqrt{64 - t^2}$ consists of all values of $t$ for which the expression under the square root, $64 - t^2$, is greater than or equal to 0. Solving the inequality $64 - t^2 \geq 0$ gives $t^2 \leq 64$, which further simplifies to $-8 \leq t \leq 8$. This means the function is defined for all $t$ in the interval $[-8, 8]$.

Find the Domain of the Function f(t)=64−t2f ( t ) = \sqrt { 64 - t ^ { 2 } }f(t)=64−t2​

Find the Domain of the Function $f ( t ) = \sqrt { 64 - t ^ { 2 } }$