WORKSHEET 26.1 Q6b

WORKSHEET 26.1 Q6c

Thx Holly

WORKSHEET 26.1 Q4a

WORKSHEET 26.1 Q3a

EX 26.1 Worksheet Q6b

Thx Camille :-)

GROVE HSC 2U EX 2.9 Q10 - not a stationary point, with asymptotes.

This function goes up to $\infty$ and down to $-\infty$ so it has no maximum or minimum value. (Grove's answer is wrong)

STATIONARY POINTS EXAMPLE WITH AN INFLEXION AS WELL

QUESTION: Consider the curve $$y=x^3-3x^2-9x+27$$
(a) Find the coordinates of any stationary points and determine their nature.
(b) Find the coordinates of any points of inflexion (and verify using a y'' box that concavity changes)
(c) Get the $x$ and $y$ intercepts. (its possible here)
(d) Sketch it!
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Before we begin just find y' and y'' right away, factorise both of them ready for use later!

$y'=3x^2-6x-9$
  $y'=3(x^2-2x-3)$
  $y'=3(x-3)(x+1)$                                                (1)

$y''=6x-6$
$y''=6(x-1)$                                                          (2)


(a) For Stat Pts: solve $y'=0$.

 $3(x-3)(x+1)=0$                                      from (1)

$\therefore x=-1, 3$

For $x=-1, y=(-1)^3-3(-1)^2-9(-1)+27=32$

For $x=3, y=(3)^3-3(3)^2-9(3)+27=0$

Stationary Points are  $(-1, 32)$  and  $(3,0)$.
Now determine their nature using a $y'$ box for each point.

For $(-1, 32)$.

 For $(3, 0)$.

[REMARK: Instead of a $y'$ box we could use $y''$ to test concavity at each $x$ value, to determine the nature. This works only if $y''\neq 0$.
Since $y''(-1)<0$ it's concave down at $x=-1$, so it's a MAX.
Since $y''(3)>0$ it's concave up at $x=3$, so it's a MIN. ]

(b)  For possible inflexion, solve $y''=0$

$6(x-1)=0$, 
$\therefore x=1$.

When $x=1$, $y=(1)^3-3(1)^2-9(1)+27=16$
Possible Inflexion is $(1,16)$.

Verify that is an inflexion using $y''$ box.

(c)  For $y$ intercept put $x=0$.  $\therefore y=27$ For $x$ intercept put $y=0$.  $\therefore 0=x^3-3x^2-9x+27$. By inspection we see that $x=-3,3$.

[REMARK: its often the case that you can't easily find the $x$ intercepts. Only do this as necessary or unless asked to do so in the question!]

(d) 

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)

GROVE HSC 2U EX 2.6 Q5 POINT OF INFLEXION

QUESTION: Show that the curve $f(x)=2x^3-5$ has an inflexion and find its coordinates.

$f'(x)=6x^2$
$f''(x)=12x$

For possible inflexions: solve $f''(x)=0$.

$12x=0$
$\therefore x=0$.
When $x=0, f(0)=2(0)^3-5=-5$.

So, possible inflexion is $(0,-5)$.

Check concavity to verify inflexion.

============================================
PS. the graph wasn't asked for but here it is anyway :-)

GROVE HSC 2U EX 2.6 Q1 - Stationary Point (Min)

Thx Holly :-)

GROVE HSC 2U EX 2.6 Q13 - stat point problem

GROVE HSC 2U EX 2.6 Q14 - stat points problem

GROVE HSC 2U EX 2.6 Q15 - PROBLEM SOLVING USING INFLEXION PROPERTY.

GROVE HSC 2U EX 2.7 Q14 with horizontal/vertical asymptote

Thx Ellie & Stephy

GROVE HSC 2U EX 2.6 Q8 stats points example using y''

Grove HSC 3U book EX 10.1 Q15 Pascal Triangle identity

Prove $\left({n\atop k}\right)=\left({n-1\atop k-1}\right)+\left({n-1\atop k}\right)$


LHS $=\left({n\atop k}\right)={n!\over k! (n-k)!}$

RHS $=\left( {n-1\atop k-1}\right)+\left({n-1 \atop k}\right)$

$={(n-1)!\over (k-1)! (n-1-(k-1))!}+ {(n-1)!\over k! (n-1-k)!}$

$={(n-1)!\over (k-1)! (n-k)!}+ {(n-1)!\over k! (n-k-1)!}$

$={k(n-1)!\over k(k-1)! (n-k)!}+ {(n-1)!(n-k)\over k! (n-k)(n-k-1)!}$

$={k(n-1)!\over k! (n-k)!}+ {(n-1)!(n-k)\over k! (n-k)!}$

$={k(n-1)! + (n-1)!(n-k)\over k! (n-k)!}$

$={(n-1)!(k + (n-k))\over k! (n-k)!}$

$={(n-1)!n\over k! (n-k)!}$

$={n!\over k! (n-k)!}$

=LHS

as required.

FACTORISE $K! + (K+1)!$

FACTORISE 

$K! + (K+1)!$

$=  K!  +   (K+1)K!$

$= K!  (1 + K + 1)$

$= K! (K+2)$

Past Prelim Exam 2011 Math Ext1 Vinnies - Q5 - Circle Geometry

Grove Prelim 3U EX 12.6, Sum & product of roots,