The function f that satisfies f'(x) = 3×2 +1, f(2)=5 is Select one: a. f(2)=23+ O…

The function f that satisfies f'(x) = 3×2 +1, f(2)=5 is Select one: a. f(2)=23+ O b. f(x) = 2 + 1 – 5 O c. f(x)=x3 ++2 d. f(x) = 3×2 – 7 O e. None of these. Let f(2) be a function that is continuous on 1, 7] with f(1) =3 and f(7) = 9. Then the Intermediate Value Theorem guarantees that Select one: O a. f'(2)=1 has at least on solution in the open interval (1,7). O b. None of these. O c. f(x) = 8 has at least one solution in the open interval (1,7). O d. f'(x) = 8 has at least on solution in the open interval (1,7). o e. f(x) attains a global maximum in the open interval (1,7).