Question

Find the relative extrema of f

Answer 1

\[f(x) = x \sqrt{4-x}\]

Answer 2

I tried but I got x = 0 and x = 4 for the critical numbers. Based on the graph, there should be one critical number around 2.5

Answer 3

Not sure where I went wrong

Answer 4

Let's think about the Domain, first.

\(4-x\ge 0 \rightarrow x\le 4\)

\(\dfrac{d}{dx}f(x) = \dfrac{-x}{2\sqrt{4-x}}+\sqrt{4-x} = \dfrac{-x}{2\sqrt{4-x}}+\dfrac{2(4-x)}{2\sqrt{4-x}} = \dfrac{8-3x}{2\sqrt{4-x}}\)

1) I'm not sure how you found x = 0 as a critical number. Please demonstrate.
2) The 1st Derivative doesn't know about the relative minimum at x = 4. How did you find that one? Please demonstrate.
3) The proper application of the derivative produces the absolute maximum at x = 8/3. Why did you NOT find this one?

Answer 5

These were my steps on finding the 1st derivative:
\[f'(x) = x(1/2)(4-x)^(-1/2) * -1 + (1)(4-x)^(1/2)\]

And I got \[f'(x) = (-1/2)(4-x)^(-1/2) + (4-x)^(1/2)\]
for the 1st derivative

Answer 6

is the 1st derivative. This might look cleaner