Find the numbers b such that the average value of f(x) = 2+6x-3x^2 on the interval [0,b] is equal to 3 I am stuck at b^2(3-b)=b and don't know what to do. Or I am doing something wrong ?
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Just solve $\displaystyle b^3-3b^2+b=0.$
Oh!!! Thanks I didnt realize that
So my answer is coming in decimals,is it ok...?
Originally Posted by racewithferrari So my answer is coming in decimals,is it ok...? I get $\displaystyle b=\frac{3\pm\sqrt{5}}{2}$.
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