Critical numbers of the function: $x\sqrt{5-x}$

Question

Let f(x) = $$\displaystyle f(x) = x\sqrt{5-x} $$

On the interval: [-6,4]

Critical numbers are the the values of x in the domain of f for which f'(x) = 0 or f'(x) is undefined.

Derivative of the function: $$ \frac{1}{2} \cdot x (5-x)^{\frac{-1}{2}} \cdot -1$$

$$ \frac {\frac{-x}{2}}{\sqrt{5-x}}$$

$f'(x) = 0$, when $x = 0, $ and is undefined when x= 5

Plugging in the roots of the derivative function and the end points of the interval into the original function: \begin{align*} f(0) & = 0\\ f(5) & = 0\\ f(-6) & = -6\sqrt{11}\\ f(4) & = 4 \cdot 1 = 4 \end{align*}

So why is the 4 not the absolute maximum value?

p.s. I assumed the first term goes to zero when taking a derivative by the product rule. I confused d/dx x = 1, with any number d/dd = 0

Answer 1

it would be much easier to work with $$y^2 = x^2(5-x), x \le 5.$$ taking the difference we find $$2y \, dy = (10x - 3x^2)\, dx $$ setting $\frac{dy}{dx} = 0$ gives you $$x = 0,\, x = \frac{10}3.$$ the spurious critical number $x = 0$ is an artifact of squaring. we know that $y = x \sqrt 5+\cdots$ for $x = 0+\cdots.$

Answer 2

Your derivative is wrong. Use the product rule to separate the $x$ from the $\sqrt{5-x}$.

$$\frac{d}{dx}\left(x\sqrt{5-x} \right) =x\frac{d}{dx}\left(\sqrt{5-x} \right)+\sqrt{5-x}\frac{d}{dx}\left(x \right)$$

$$=x\frac{1}{2\sqrt{5-x}}(-1)+\sqrt{5-x}\cdot 1$$ $$=\frac{-x}{2\sqrt{5-x}}+\frac{2(5-x)}{2\sqrt{5-x}}$$ $$=\frac{10-3x}{2\sqrt{5-x}}$$

Setting this equal to zero, you get

$$x=\frac{10}3$$

This is undefined only at $x\ge 5$, which is not in your interval of interest. You should be able to continue from here.

Answer 3

$\dfrac{d(x\sqrt{5-x})}{dx}=\dfrac{10-3x}{2\sqrt{5-x}}$

$10-3x=0$ imply the critical number is $x=\dfrac{10}{3}$.