Asked by anonymous
approximate all zeros and extrema to the nearest hundredth. how many zeros, if any, are imaginary.
-x^4+2x^2+x-4.
Are there no zeros?
-x^4+2x^2+x-4.
Are there no zeros?
Answers
Answered by
Steve
not sure what techniques you have to use. You can use Descartes' Rule of Signs, or the signs produced in synthetic division to get a handle on the bounds, but otherwise you may be stuck with iterative numeric techniques.
If you have calculus under your belt, then we can check for extrema and concavity:
y = -x^4+2x^2+x-4
y' = -4x^3+4x+1
y" = -12x^2+4
Now we're not much better off. The derivatives also have no rational zeros. You can check y' and find that it is zero at x = -.837, -.269, 1.107
at hose values of x, y= -3.93, -4.13, -1.94
also, y" is negative, positive, negative
That means that we have a global maximum at x=1.107, and y is negative there, so there are no possible roots for y.
If you have calculus under your belt, then we can check for extrema and concavity:
y = -x^4+2x^2+x-4
y' = -4x^3+4x+1
y" = -12x^2+4
Now we're not much better off. The derivatives also have no rational zeros. You can check y' and find that it is zero at x = -.837, -.269, 1.107
at hose values of x, y= -3.93, -4.13, -1.94
also, y" is negative, positive, negative
That means that we have a global maximum at x=1.107, and y is negative there, so there are no possible roots for y.
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