Curve Sketching with $y''=0$ but NO inflexion.

Curve Sketching with $y''=0$ but NO inflexion.

Consider the curve, 
      $$y=x^4-8x^3+24x^2-288x+5$$

      $y'=4x^3-24x^2+48x-288$

      $y'=4x^2(x-6)+48(x-6)$
      $y'=(4x^2+48)(x-6)$

      $y'=4(x^2+12)(x-6)$

      $y''=12x^2-48x+48$

      $y''=12(x^2-4x+4)$

      $y''=12(x-2)^2$

Stationary Points: Solve $y'=0$.
$\therefore x=6$ and so the stationary point is $(6,-1291)$.

When $x=6$, $y''=12(6-2)^2>0$ and hence it's a minimum turning point. 

Possible Inflexions: Solve $y''=0$.
$\therefore x=2$ and the possible inflexion point is $(2, -523)$.

Check the sign of $y''$ either side of the possible inflexion. 

$x$	$1$	$2$	$3$
$y’’$	$+$	$0$	$+$

The sign of $y''$ stays positive, and so this is NOT an inflexion.
The curve just flattens out at the point where $x=2$. See graph.



Here are a list of curves with similar features, in that y''=0 but not inflexion!

• $y=x^4-8x^3+24x^2-288x+5$ (above)

• $y=3x^4-12x^3+18x^2-108x+1$

• $y=x^4-4x^3+6x^2-36x-10$

• $y=3x^4-8x^3+8x^2-32x+11$

• $y=3x^4-20x^3+50x^2-500x$

• $y=3x^4-4x^3+2x^2-56x+2$ ($y'$ has a $x-2$ as a factor, hint :-))

Question: (Calculus – curve sketching – polynomials) Sketch the curve given by $y = x(4− x^2)$ . Clearly indicate the stationary points and all the intercepts. (so find the coordinates of any stationary points)

Question: (Calculus – curve sketching – polynomials)
Sketch the curve given by $y = x(4− x^2)$ . Clearly indicate the stationary points and all the intercepts. (so find the coordinates of any stationary points)

GROVE HSC 3U EX 2.1 Q14/15

GROVE HSC CHALLENGE EX2 - Q2 - Curve sketching, stat points, inflexions

Thanx Bec Rae.

GROVE HSC 2U/3U PRACTICE ASSESSMENT TASK 1 - Q43 Curve Sketching

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 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 Q8 stats points example using y''