To find the critical points of the function [tex]\( y = 3x \sqrt{36 - x^2} \)[/tex], we need to follow these steps:
1. Determine the Domain:
[tex]\[
y = 3x \sqrt{36 - x^2}
\][/tex]
The expression under the square root must be non-negative:
[tex]\[
36 - x^2 \geq 0 \implies -6 \leq x \leq 6
\][/tex]
So, the domain of [tex]\( y \)[/tex] is [tex]\( [-6, 6] \)[/tex].
2. Find the First Derivative:
[tex]\[
y = 3x \sqrt{36 - x^2}
\][/tex]
To find the critical points, we need to take the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
y' = \frac{d}{dx} \left( 3x \sqrt{36 - x^2} \right)
\][/tex]
Using the product rule and chain rule, we get:
[tex]\[
y' = 3 \sqrt{36 - x^2} + 3x \cdot \frac{d}{dx} \left( \sqrt{36 - x^2} \right)
\][/tex]
[tex]\[
y' = 3 \sqrt{36 - x^2} + 3x \left( \frac{1}{2} (36 - x^2)^{-1/2} \cdot (-2x) \right)
\][/tex]
Simplifying:
[tex]\[
y' = 3 \sqrt{36 - x^2} - \frac{3x^2}{\sqrt{36 - x^2}}
\][/tex]
Combining the terms over a common denominator:
[tex]\[
y' = \frac{3 (36 - x^2) - 3x^2}{\sqrt{36 - x^2}} = \frac{3(36 - 2x^2)}{\sqrt{36 - x^2}} = \frac{108 - 6x^2}{\sqrt{36 - x^2}}
\][/tex]
3. Set the First Derivative to Zero:
Set [tex]\( y' = 0 \)[/tex]:
[tex]\[
\frac{108 - 6x^2}{\sqrt{36 - x^2}} = 0
\][/tex]
The numerator must be zero for the fraction to be zero:
[tex]\[
108 - 6x^2 = 0 \implies 6x^2 = 108 \implies x^2 = 18 \implies x = \pm 3\sqrt{2}
\][/tex]
4. Identify and List the Critical Points:
The critical points occur at:
[tex]\[
x = -3\sqrt{2} \quad \text{and} \quad x = 3\sqrt{2}
\][/tex]
So, the critical points of the function where the first derivative is 0 are:
[tex]\[
(-3\sqrt{2}, 3\sqrt{2})
\][/tex]
Thus, the correct choice is:
A. The critical point(s) where [tex]\( f^{\prime} \)[/tex] is 0 is/are at [tex]\( x = -3\sqrt{2}, 3\sqrt{2} \)[/tex].
