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Saturday, 15 July 2017

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 :-))



Thursday, 29 June 2017

Curve Sketching y=x^x

\begin{array}{ll} x^2 &=9\\ x &=\pm 3 \end{array}

Harder example: (stationary points and curve sketching.
Consider the curve
y=x^x, \quad x>0
Find the coordinates of any stationary points and inflexions.
Sketch the curve.
[You may assume that \lim_{x\rightarrow 0} x^x=1. This can be proven by using logs and l’Hopital’s rule.]

See for example, 



Answers

y'=x^x(\ln x+1)
y''=x^x[(\ln x+1)^2+{1\over x}]
Stationary Point is \left( 1/e,(1/e)^{1/e} \right) = approx (0.37,0.69) = approx (0.4,0.7) Minimum

Also note that y''>0 for x>0





Thursday, 23 March 2017

Given that \alpha, \beta are the roots of the equation 3x^2+2x+7=0, find the equation of the quadratic equation with roots \alpha^2, \beta^2.

 Given that \alpha, \beta are the roots of the equation 3x^2+2x+7=0, find the equation of the quadratic equation with roots \alpha^2, \beta^2.