GROVE HSC 2U EX 2.6 Q11 STATIONARY POINT - TRICKY FUNCTION

QUESTION: Find any stationary points on the curve $y=(x-3)\sqrt{4-x}$

This is a product of two functions, so we must use the Product rule to differentiate it.

If $y=uv$, $$y'=vu'+uv'$$

$u=x-3$         $u'=1$

$v=\sqrt{4-x}$
$v=(4-x)^{1\over 2}$
$v'={1\over 2}(4-x)^{-{1\over 2}}(-1)$
$v'=\displaystyle {-1\over 2\sqrt{4-x}}$

[RECALL chain rule. If $y=f(x)^n$, then $$y'=n\,f(x)^{n-1}\times f'(x)$$ ]

Therefore,
$y'=vu'+uv'$
$y'=\sqrt{4-x} .1 + (x-3)\left( \displaystyle {-1\over 2\sqrt{4-x}}\right)$

$y'=\displaystyle {\sqrt{4-x} \over 1} + \displaystyle{-(x-3) \over 2\sqrt{4-x}}$

$y'=\displaystyle {\sqrt{4-x} \over 1} + \displaystyle{3-x \over 2\sqrt{4-x}}$

$y'=\displaystyle {\sqrt{4-x} \over 1}\times {2\sqrt{4-x} \over 2\sqrt{4-x}} + \displaystyle{3-x \over 2\sqrt{4-x}}$

[RECALL: $\sqrt{a}\times \sqrt{a}=a$]

$y'=\displaystyle{2(4-x)+3-x \over 2\sqrt{4-x}}$

$y'=\displaystyle{8-2x+3-x \over 2\sqrt{4-x}}$

$y'=\displaystyle{11-3x\over 2\sqrt{4-x}}$

FOR STAT POINTS: SOLVE $y'=0$.

$\displaystyle{11-3x\over 2\sqrt{4-x}}=0$

$\therefore 11-3x=0$

$3x=11$

$x=\displaystyle {11\over 3}$

When
$x=\displaystyle {11\over 3}$,
$y=\displaystyle\left({11\over 3}-3\right)\sqrt{4-{11\over 3}}$
$y=\displaystyle{2\over 3}\sqrt{{1\over 3}}$
$y=\displaystyle{2\over 3\sqrt{3}}$

$y=\displaystyle{2\sqrt{3}\over 9}$

Stationary point is $\displaystyle\left({11\over 3},{2\sqrt{3}\over 9}\right)$.

Now determine the nature of this stat point.

So it's a maximum turning point.

Correct to 2 decimal places it's $(3.67, 0.38)$. (Grove gives the answer to 2 d.p. even though the question didn't ask for it)