Find $a$, such that the function $f ( x ) = 4 x + \sqrt { a - x ^ { 2 } }$ has the domain $( - 4,4 )$.
Select the correct answer.
A) $a = - 16$
B) $a = \sqrt { 4 }$
C) $a = - \sqrt { 4 }$
D) $a = 16$
E) $a = 4$

Question

Quizplus · Accepted Answer

To find the domain of the function, we need to ensure that the term inside the square root is non-negative. That is, we need to have
$$a-x^2\geq 0$$
This inequality holds for all $x$ in $(-4, 4)$ if and only if the maximum value of $x^2$ over this interval does not exceed $a$. Now, note that $(4)^2=16$ is the maximum value that $x^2$ can take in the given interval. Therefore, we must have
$$a\geq 16$$
In order to ensure that $f(x)$ is defined for all $x$ in $(-4,4)$, we also need to make sure that the expression $\sqrt{a-x^2}$ is non-negative. This means we need to have
$$a-x^2\geq 0\Rightarrow \sqrt{a-x^2}	ext{ is defined}\quad	ext{ for all } x\in (-4,4)$$
In other words, we need the minimum value of $a-x^2$ over the interval $(-4,4)$ to be non-negative. Now, note that $(-4)^2=16$ is the maximum value that $-x^2$ can take in the given interval. Therefore, we want to look at the minimum value that $a-x^2$ can take on, which occurs at $x=0$. Thus, we must have
$$a-0^2\geq 0\Rightarrow a\geq 0$$
Putting these inequalities together, we get $a\geq 16$ and $a\geq 0$. The smallest possible value of $a$ that satisfies both these inequalities is $a=16$. Thus, the answer is $\boxed{	extbf{(D)}}$.

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Find $a$ , Such That the Function $f ( x ) = 4 x + \sqrt { a - x ^ { 2 } }$