Question

4. [-/1 Points] DETAILS LARCALC11 3.2.006. Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that fe 8(+) = N(2 – 2+)'. 1.1,1) None of these. f(a) does not equal f(b) for all possible values of a and b in the interval [-1, 1). Fla) does not equal f(b) for any values in the interval [-1, 1). There are points on the interval [a, b] where fis not continuous. There are points on the interval (a, b) where f is not differentiable.

Answer 1

f(x) = 12-2')> ,[1] f(1) = 82-C-133)* 3e-1)3 = 1 fc1) 12-(1923) hance f(-1) = f(1) √(2-13 = 1 Now f(x) = √(2-x73) will be defined for (2.-2433 0 2-743 20 ) sol (x/3² <02 =) - 52 < x 282 (-2) ² < x < (123 - - 212 ca<282 hence tex) is continuous in E!!]

Now f(x) = √ (2 - x733 √(2 – 2 73}} = (2- 233% fw-402-2)*(0-12" a f'(X) = (2-3*(-3x+3) = f'(x) = كما زيلا -(2-x²3) x f'(x) will not be defined at x2:00 Thus Cast option that is There are some points on the interval (a, b) where of is not differenbable is correct.