Question

The maximum and minimum values of the function y = x3 - 3x2 +6 what ia the procedure
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Answer 1

Answer:

y=x³-3x²+6

dy/dx=3x²-6x

for max or min values,

dy/dx =0

3x²-6x=0

x(3x-6)=0

x=0,2

d²y/dx²=6x-6

let x=0,

d²y/dx²=-6(-ve)

so, fx is max at x=0

let, x=2,

d²y/dx²=6(+ve)

so, fx is min at x=2,

maximum value=x³-3x²+6

put x=0,

max value=6

min value,

put x=2,

min value=8-12+6=2