There's
$$y=(x-1)^2(x-2)^3$$
I got confused due to post derivative simplification, tried to take it 4 times :/
How to find minima and maxima of $y=(x-1)^2(x-2)^3$?
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$\begingroup$
derivatives
maxima-minima
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0differentiate with the product rule? -then you will see that (x-1) and (x-2) are also factors of the derivative - does that speed it up? you can then factor them out – 2017-02-21
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0@satyatech I'm pretty sure that the maximum of $-x^2$ is not $\infty$ :p – 2017-02-21
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0@Zubzub put x= - infinity in the function (in the qn) you will get -infinity .And for your Question the of $$\ max(-x^2) =0 $$ – 2017-02-21
1 Answers
2
We get: $y'(x)=2(x-1)(x-2)^3+3(x-2)^2(x-1)^2$.
Is this useful ?