Answer to Question #17684 in Calculus for hsd
defined on the interval [−5,7].
a.) f(x) is concave down on the interval .
b.) f(x) is concave up on the interval .
c.) The minimum for this function occurs at x= .
d.) The maximum for this function occurs at x= .
x0 where f(x) does not exist and f(x)->infinity,x->x0 so x^2-1=0 then x=1, x=-1
b) we must find f"
f"=0 then x=0
f"(2)>0 but f"(1/2)<0
so f(x) cant be concave up on the given region
c) points of reflection f"=0 so x=0
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