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What are the real zeros for f(x)=x^3-3x^2+8
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yes please
Well, there is only one zero here which is x=-1.3553
You can get this if you take the derivative of the equation.
*one real zero.
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Can you draw or show how you solved it please I'm not sure as to how you take the derivative of the equation?
I don't want to write out all the steps.
But, I can paste an image of a computer solution. If you want.
Ok that'll be good thanks!
It will show all the steps.
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Okay thanks!
That should help!
BTW, if you want to solve something like this in the future, you may want to look into the Durand-Kerner Method. Very few students know about this.
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