Use the Intermediate Value Theorem to show that there is a root of the equation m3 m2 + 2m 1 = 0 on the interval (0, 1). Let f(a:) = C] . Since f(9:) is a function, it is continuous for all real number. In particular, it is continuous on the interval C] . Since N z [j is between f(0) = C] and f(1) = [:], and f(0) f(1), there exist a least one number 0 in the interval C] such that f(c) = C]. Therefore, by the Intermediate Value Theorem there exist a root to the equation 3:3 m2 + 2:1: 1 = 0 on the interval (0, 1). Find the constants, a, b, c, and d for f(x) = asin(b(a: c)) + d such that f(a:) would have to -" following graph. Find the following limits, if the limit does not exist input DNE for the answer. sin(0) a. lim 0-0 3x + 1 b. lim x-2 x -4 c. lim [x] x-> 2 a + 4 d. lim x-7 -4+ a + 4 2 + 4 e. lim = x7-4- x +4 ac + 4 f. lim4 Find lim . t>2 t3 8 With direct substitution, you get C] over C] , which mean you need to do some algebra. t2 By factoring t3 8 and simplifying, you have... 94 lim = lim t>2 t3 _ 8 t>2 Now with direct substitution, you have