Given
f(x)=x4−8x2+16
Since
f′(x)=4x3−16xf′′(x)=12x2−16
Solve
f′′(x)=0
we get
x=323, x=−323
Now , we check the sign of f′′(x) , since
f′′(−2)>0, f′′(0)<0, f′′(2)>0
Then f(x) is concave up on (−∞,−23/3)∪(23/3,∞) and
concave down on (−23/3,23/3)
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