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Determine the critical numbers, if any, of the function f f on the interval [1,8] [1,8] .\newlinef(x)=x28x f(x)=x^{2}\sqrt{8-x} \newlineGive your answer as a comma-separated list. Express numbers in exact form. If the function does not have any critical numbers, enter DNE.

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Q. Determine the critical numbers, if any, of the function f f on the interval [1,8] [1,8] .\newlinef(x)=x28x f(x)=x^{2}\sqrt{8-x} \newlineGive your answer as a comma-separated list. Express numbers in exact form. If the function does not have any critical numbers, enter DNE.
  1. Identify Critical Numbers: Identify the critical numbers of the function f(x)=x28xf(x) = x^2 \sqrt{8-x}. Critical numbers occur where the derivative f(x)f'(x) is zero or undefined.
  2. Calculate Derivative: Calculate the derivative of f(x)=x28xf(x) = x^2 \sqrt{8-x}. Use the product rule and the chain rule.\newlinef(x)=ddx[x2]8x+x2ddx[8x]f'(x) = \frac{d}{dx} [x^2] \sqrt{8-x} + x^2 \frac{d}{dx} [\sqrt{8-x}]\newlinef(x)=2x8x+x2(12)(8x)12(1)f'(x) = 2x \sqrt{8-x} + x^2 \left(\frac{1}{2}\right) \left(8-x\right)^{-\frac{1}{2}} \left(-1\right)\newlinef(x)=2x8x(x228x)f'(x) = 2x \sqrt{8-x} - \left(\frac{x^2}{2 \sqrt{8-x}}\right)
  3. Simplify Derivative: Simplify the derivative to find where it is zero or undefined.\newlinef(x)=2x(8x)x228xf'(x) = \frac{2x \cdot (8-x) - x^2}{2 \cdot \sqrt{8-x}}\newlinef(x)=16x2x2x228xf'(x) = \frac{16x - 2x^2 - x^2}{2 \cdot \sqrt{8-x}}\newlinef(x)=16x3x228xf'(x) = \frac{16x - 3x^2}{2 \cdot \sqrt{8-x}}
  4. Find Zeroes: Set the numerator equal to zero to find where the derivative is zero.\newline16x3x2=016x - 3x^2 = 0\newlinex(163x)=0x(16 - 3x) = 0\newlinex=0x = 0 or x=163x = \frac{16}{3}
  5. Check Values: Check if the values found are in the interval [1,8][1,8].x=0x = 0 is not in the interval [1,8][1,8].x=163x = \frac{16}{3} is approximately 5.335.33, which is in the interval [1,8][1,8].
  6. Check Undefined Points: Check for points where the derivative is undefined by setting the denominator equal to zero.\newline28x=02 \cdot \sqrt{8-x} = 0\newline8x=0\sqrt{8-x} = 0\newline8x=08 - x = 0\newlinex=8x = 8
  7. Verify Endpoint: Verify that x=8x = 8 is in the interval [1,8][1,8]. Since 88 is the endpoint of the interval, it is considered a critical number if the function is defined there.
  8. Combine Critical Numbers: Combine the results from steps 55 and 77 to list all critical numbers in the interval [1,8][1,8].\newlineThe critical numbers are x=163x = \frac{16}{3} and x=8x = 8.

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