Questions: Determine the critical numbers, if any, of the function (f) on the interval ([1,3]). (f(x)=x^2 sqrt3-x) Give your answer as a comma-separated list. Express numbers in exact form. If the function does not have any critical numbers, enter DNE. (x=)

Solution Steps

To find the critical numbers of the function \( f(x) = x^2 \sqrt{3 - x} \), we first compute its derivative using the product rule. The derivative is given by:

\[ f'(x) = -\frac{x^2}{2\sqrt{3 - x}} + 2x\sqrt{3 - x} \]

Next, we set the derivative equal to zero to find the critical points:

\[ -\frac{x^2}{2\sqrt{3 - x}} + 2x\sqrt{3 - x} = 0 \]

This simplifies to:

\[ 2x\sqrt{3 - x} = \frac{x^2}{2\sqrt{3 - x}} \]

Multiplying both sides by \( 2\sqrt{3 - x} \) (assuming \( 3 - x \neq 0 \)) leads to:

\[ 4x(3 - x) = x^2 \]

Rearranging gives:

\[ 4x(3 - x) - x^2 = 0 \implies 12x - 5x^2 = 0 \]

Factoring out \( x \):

\[ x(12 - 5x) = 0 \]

Thus, the critical points are:

\[ x = 0 \quad \text{and} \quad x = \frac{12}{5} \]

We need to check which of these critical points lie within the interval \([1, 3]\):

• \( x = 0 \) is not in the interval.
• \( x = \frac{12}{5} = 2.4 \) is within the interval.

Final Answer