Posted by Salman on Sunday, November 22, 2009 at 9:22am.
To find the minimum of the function, you need to find the derivative, f'(x). At the local minimums, the derivative equals zero. That gives you two of the three points, and for which you can solve using the quadratic formula.
Since the function is not a polynomial (because of the square-root function), its domain is not (-∞,∞).
You will need to find the domain of the function, namely the interval of which the expression under the square-root sign stays positive. The said expression is a cubic and crosses the x-axis only once, so it is easy to find the zero using Newton-Raphson method. In case you have difficulties, it is around -43.
Additional note:
do not forget to test the first two critical points (where f'(x)=0) for maximum or minimum.
what is the exact value of the -43 term? i cant find it
Tell me what values you've got so far.
It should be quite straight forward, because the slope is almost uniform in that stretch.
Perhaps you didn't get the right derivative, or formula.
Oh, it's not a factoring process, the third point is to be solved using Newton-Raphson. The first two are the solutions of the quadratic.
I cant find the value of x1. im getting it around -43.2423, but this answer is not correct
Mine agrees with yours up to -43.24...
To how many figures are you supposed to calculte, and how many lives do you have left (assuming you have a computer exercise).