The point where the derivative vanishes are the roots of the equation and the only real root can be easily found with the Newton Raphson iterations ... for the second derivative is so that the point is a minimum...
Function f(x)=x^4+8x^2-48x+19 has one extreme value for a x-value in 1<x<2 interval. Decide this value with 2 decimals and if that is max/min.
I would be grategul if any body could help me.
The Newton Raphson iterations are useful for the search of a root of an equation like . The procedure obtain an more and more precise approximation...
... starting from a given value . In Your case is...
... so that the iterations produce...
... and for the result pratically doesn't change...
I would suggest that being given an interval like this indicates that may-be the student is expected to use the bisection algorithm for this problem.
Using Newton's algorithm 3 iterations allow three digits precision and 5 iteration allow a twelve digits precision. With the bisection algorithm for the same precision You need 9 and 38 iterations respectively... it seems to be a very time and work spending approach...
By the way, "extreme values are always found where the derivative of the function equals zero" is NOT true. On an interval that is not closed and bounded (does not have or does not contain its endpoints) there may be NO extreme values. On an interval that is closed and bounded, extreme values will occur at one of three different kinds of points:
1) Where the derivative is 0.
2) Where the derivative does not exist.
3) At an endpoint of the interval.
Here, the given function is a polynomial so the derivative always exists. And the interval is not closed so does not contain its endpoints so, in this particular example, you do end up with just (1).
There is a bit more fiddling involved with NR to get the same extras (and more work as we need derivatives).