# Math Help - Finding extreme value of a fourth degree function in a given interval.

1. ## Finding extreme value of a fourth degree function in a given interval.

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.

2. The point where the derivative vanishes are the roots of the equation $x^{3} + 4 x -12 =0$ and the only real root can be easily found with the Newton Raphson iterations ... for $1< x< 2$ the second derivative is $>0$ so that the point is a minimum...

Kind regards

$\chi$ $\sigma$

3. is x=2 an possible answer because $f'(2) \neq 0$

4. But $x=2$ is not a root of $x^{3} +4 x -12 =0$ ...

Kind regards

$\chi$ $\sigma$

5. So what will be the answer?

6. The Newton Raphson iterations are useful for the search of a root of an equation like $f(x)=0$. The procedure obtain an more and more precise approximation...

$\displaystyle x_{n+1} = x_{n} - \frac{f(x_{n})}{f^{'} (x_{n})}$ (1)

... starting from a given value $x_{0}$. In Your case is...

$f(x)= x^{3} + 4\ x -12 \implies f^{'}(x)= 3\ x^{2} + 4$

... so that the iterations produce...

$x_{0}= 1$

$x_{1}= 2$

$x_{2}= 1.75$

$x_{3}= 1.72274881517...$

$x_{4}= 1.72244823567...$

$x_{5}= 1.72244819948...$

... and for $n>5$ the result pratically doesn't change...

Kind regards

$\chi$ $\sigma$

7. No, x= 2 is not a possible answer, but not because of the fact that it does not make the derivative 0. And endpoint of a closed interval can be a max or min without the derivative being 0 there. However, "2" is NOT in the interval you gave, (1, 2).

8. From what I remember being taught the extreme values are always found where the derivative of the function equals zero, since it means the slope there is zero. I do not understand that long and complicate post. Why is $x_{0}= 1$ a given value?

9. Originally Posted by BugzLooney
... why is $x_{0}= 1$ a given value?...
In general the convergence of the Newton Raphson iteration is not guaranted for all the possible value of $x_{0}$. For example the [trivial] second order equation $x^{2}-1=0$ has roots $\pm 1$ but is $f^{'} (0)=0$ so that starting with $x_{0}$ is not the best possible way ... In our case however is $f^{'} (x)>0$ for all x and any value of $x_{0}$ is valid...

Originally Posted by BugzLooney
... I do not understand that long and complicate post...
May be that someone strongly aspires to became 'Best Mathematician' and thinks that the best way to meet the goal is to exaltate the 'theoretical aspects' of the problems... that isn't the only post where that happens ...

Kind regards

$\chi$ $\sigma$

10. Originally Posted by chisigma
The point where the derivative vanishes are the roots of the equation $x^{3} + 4 x -12 =0$ and the only real root can be easily found with the Newton Raphson iterations ... for $1< x< 2$ the second derivative is $>0$ so that the point is a minimum...

Kind regards

$\chi$ $\sigma$
Since f'(x) is -ve at 1 and +ve at 2 the bisection algorithm can be used on the derivative to locate a root in [1,2].

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.

CB

11. Originally Posted by CaptainBlack
Since f'(x) is -ve at 1 and +ve at 2 the bisection algorithm can be used on the derivative to locate a root in [1,2].

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.

CB
It is very interesting to compare the 'efficiency' of the bisection and Newton's algorithms for the search of the root between 1 and 2 of the equation $\displaystyle x^{3} + 4\ x -12=0$. The iteration for the Newton Raphson algorithm are reported in my previous post. The first iterations of the bisection algorithm are...

$\overrightarrow{x}_{0}= [1,2]$

$\overrightarrow{x}_{1}= [1.5,2]$

$\overrightarrow{x}_{2}= [1.5,1.75]$

$\overrightarrow{x}_{3}= [1.625,1.75]$

$\overrightarrow{x}_{4}= [1.6875,1.75]$

$\overrightarrow{x}_{5}= [1.71875,1.75]$

$\overrightarrow{x}_{6}= [1.71875,1.734375]$

$\overrightarrow{x}_{7}= [1.71875,1.7265625]$

$\overrightarrow{x}_{8}= [1.71875,1.72265625]$

$\overrightarrow{x}_{9}= [1.720703125,1.72265625]$

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...

Kind regards

$\chi$ $\sigma$

12. Originally Posted by BugzLooney
From what I remember being taught the extreme values are always found where the derivative of the function equals zero, since it means the slope there is zero. I do not understand that long and complicate post. Why is $x_{0}= 1$ a given value?
Chi Sigma is not suggesting that x= 1 may be a solution. But to use Newton's method, it helps to start from a point that you know is close to a solution. "1" is reasonably close to any number in (1, 2). Chi Sigma could have used 2 or any number in (1, 2) as a starting value. I imagine he used "1" because it seemed to give simpler intermediate values.

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).

13. Originally Posted by chisigma
It is very interesting to compare the 'efficiency' of the bisection and Newton's algorithms for the search of the root between 1 and 2 of the equation $\displaystyle x^{3} + 4\ x -12=0$. The iteration for the Newton Raphson algorithm are reported in my previous post. The first iterations of the bisection algorithm are...

$\overrightarrow{x}_{0}= [1,2]$

$\overrightarrow{x}_{1}= [1.5,2]$

$\overrightarrow{x}_{2}= [1.5,1.75]$

$\overrightarrow{x}_{3}= [1.625,1.75]$

$\overrightarrow{x}_{4}= [1.6875,1.75]$

$\overrightarrow{x}_{5}= [1.71875,1.75]$

$\overrightarrow{x}_{6}= [1.71875,1.734375]$

$\overrightarrow{x}_{7}= [1.71875,1.7265625]$

$\overrightarrow{x}_{8}= [1.71875,1.72265625]$

$\overrightarrow{x}_{9}= [1.720703125,1.72265625]$

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...

Kind regards

$\chi$ $\sigma$
The difference is that for a continuous function the bisection method is guaranteed to find a root (to what ever precision we require) given the product of the function values at each end of the initial interval is negative, also at each step we have an interval containing the root.

There is a bit more fiddling involved with NR to get the same extras (and more work as we need derivatives).

CB