1. Let f(x)=√(x - 1)² with domain [a, b], where a = -1 and b = 3.(a) Is f(x) continuous on its domain (continuous at every point of its domain)?(b) Is f(x) differentiable at each point of its domain? If not, make a list of the points in (-a, b) wheref(x) is not differentiable.Such points are not in the domain of f'(x).(c) Find f(a) and f(b)f(b) f(a),=-(d) Is there a number c in (a, b) such that f'(c)(e) Which hypothesis of Rolle's theorem is not satisfied?b-aYes or No 2. Sketch the graph of a continuous function f(x) with domain [-1,8] which is differentiable on (-1,8)and satisfies:(a) f'(x) > 0 in (-1, 1) and (3,6).(b) f'(x) < 0 in (1,3) and (6,8).(c) f(x) is concave down in (-1,2) and (5,8)(d) f(x) is concave up in (2, 5).For your function, list (as sets) the points where f(x) attains local maxima, and where it attains localminima.3212468

Question (1) f(x) = (x-1)^25 with domain [-1, 8] (a) Checking continuity, At x = -1, Since it's the left end of the domain so for f(x) to be continuous here, f (-1) should be equal to limh->0 f (-1 +h) f (-1) = (-1-1)^25 =45 limh->0f (-1+h) = limh->0(-1 + h-1)^25 = limh->0h^2 + 4 - 4h5 Putting h = 0, limh->0f (-1+h) = 45 So, the function is continuous at x = -1. Similarly, at x = 8, Since it's the right end of the domain so for f(x) to be continuous here, f (8) should be equal to limh->0 f (8 - h) f (8) = (8-1)^25 = 495 limh->0 f (8 - h) =limh->0(8-h-1)^25 =limh->0 h^2 + 49 - 14h5 Putting h = 0, limh->0 f (8 - h) = 495 Since the function is continuous at both ends of the domain and there is no value of x in the domain where the function is discontinuous, therefore, the function is continuous at the given domain. (b) Checking differentiability, f'(x) = 2/ (5(x - 1)^35) Since the denominator cannot be zero, x - 1 is not equal to 0 x is not equal to 1. x = 1 lies at the domain of the function Therefore, the function is not differentiable at x = 1. (c) f(a) = f (-1) = (-1-1)^25 =45 f(b) = f (8) = (8-1)^25 = 495 (d) The mean value theorem states that if a function f is defined on the closed interval [a,b] and is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then there exists a value x = c in such a way that f'(c) = [f(b) – f(a)]/(b-a) But since the function is not differentiable on the open interval (a, b) therefore this function doesn't have a point x=c, such that f'(c) = [f(b) – f(a)]/(b-a) (e) Rolles's theorem is a special case of the 'mean value of theorem' where the condition is f(a) = f(b) and f'(c) = 0. In this function a is not equal to b so that hypothesis itself is wrong. Question (2) We are given the sign chart of f'(x), So, we get to see the critical points of the function which are, x = 1, 3, and 6 at these points f'(x) = 0. At x = 1 the sign of f'(x) changes from +ve to -ve that means at x = 1 lies the local maximum. At x = 3 the sign of f'(x) changes from -ve to +ve that means at x = 3 lies the local minimum. At x = 6 the sign of f'(x) changes from +ve to -ve that means at x = 6 lies the local maximum. Therefore, at (1, f (1)) and (6, f (6)) there exist local maxima and at (3, f (3)) there exists local minima. We also know that x = 2 and x = 5 the function changes the nature of its curvature which means here at these two values of x there are two inflection points.