Question

The radius of the interval [-1,3] is​

Answer 1

Answer:

ANSWER

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]