The radius of the interval [-1,3] is
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f
′
(x)=12(x
2
−1)
f
′
(x)=0 at x=±1
f
′
(x)<0 If ∣x∣<1
f
′
(x)>0 If ∣x∣>1
f(x) is min when x=1
f(1)=−8
f(x) is max either at x=−1 or x=3
f(−1)=8 , f(3)=72
So image [-8, 72] in [-1, 3]
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