maxium value of
х^3-18 х^2+96x
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Answered by
7
Answer:
160
Step-by-step explanation:
f(x)=x3−18x2+96x
f′(x)=3x2−36x+96
f′(x)=x2−12x+32=0
=x2−8x−4x+32=0
=x(x−8)−4(x−8)
f′(x)=(x−4)(x−8)=0
x=4,8
f′(x)=2x−12
f′(4)=−4<0
x=4 point of maxima
f(4)=64−16×18+96×4
=448−288=160
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