# Find the point of

local max. value is 68 at x = 1 and local min. values are 1647 at x = 6 and 316 at x = 5

F’(x)=4x3-124x+120=0

4(x3-31x+30)=0

For x=1, the given eq is 0

؞x-1 is a factor,

4(x-1)(x+6)(x-5)=0

X=1,-6,5

F’’(1)<0, 1is the point of max.

F’’(-6) and f’’(5)>0, -6 and 5 are point of min.

F(1)=68

F(-6)=-1647

F(5)=-316

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