Use the Newton's method to find the inflexion points of a function given by (x^3+3x^2-4)/e^x

The derivative of the function is -(x^3-6x-4)/e^x

The Newton's method has the formula xn+1=xn-(f(xn))/f'(xn)

I am unable to proceed with x0 values of 1 and -2.

What initial value should I use?Please help me.


Hint....To find points of inflexion you need to solve $\frac{d^2y}{dx^2}=0$

  • $\begingroup$ I know that.....but how do you do it using Newton's method? $\endgroup$ – Amarjeet Jayanthi Aug 6 '15 at 14:44
  • $\begingroup$ Do you have an expression for the second derivative? $\endgroup$ – David Quinn Aug 6 '15 at 14:47
  • $\begingroup$ The second derivative is (x^3-3x^2-6x+2)/e^x. $\endgroup$ – Amarjeet Jayanthi Aug 6 '15 at 14:49
  • $\begingroup$ So set $f(x)=$ the top line of this and apply Newton's Method. There are three possible points. Which one do you need to find? $\endgroup$ – David Quinn Aug 6 '15 at 14:51
  • $\begingroup$ I need to find all of them.How did you get them and what are they? $\endgroup$ – Amarjeet Jayanthi Aug 6 '15 at 14:53

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