calculus problems

For a particular Critical value x = a, find f " ' (a) (i) If f ''(a) < 0 then f (x) has a local maxima at x = a and f (a) is the maximum value. (ii) If f ''(a) > 0 then f (x) has a local minima at x = a and f (a) is the minimum value. (iii) If f ''(a) = 0 or , the test fails and the first d..

f '(x) = 6x 2 - 42x + 36 f '(x) = 0 x = 1 and x = 6 are the critical values f ''(x) =12x - 42 If x =1, f ''(1) =12 - 42 = - 30 < 0 x =1 is a point of local maxima of f (x). Maximum value = 2(1) 3 - 21(1) 2 + 36(1) - 20 = -3 If x = 6, f ''(6) = 72 - 42 = 30 > 0 x = 6 is a point..

Let f be a continuous function on an interval I = [a, b]. Then, f has the absolute maximum value and f attains it at least once in I. Also, f has the absolute minimum value and attains it at least once in ..

Find all the points where f ' takes the value zer..

At all the points calculate the values of ..

If a quantity y varies with respect to another quantity 'x' satisfying some rule y = f(x), in other words if y is a function x, then represents the rate of change of y with respect to x For x = x 0 , dy/dx at x 0 is called the rate of change of y with respect to x at x 0 . If y is a function of t x..

Let f be a real valued function in [a,b] such that f is continuous in [a,b]. f is differentiable in (a,b..

We cannot obtain c if any one of the conditions of Rolle's theorem are not satisfie..

Verify Rolle's theorem for the function f (x) = x 2 - 8x + 12 on (2, 6) Since a polynomial function is continuous and differentiable everywhere f (x) is differentiable and continuous (i) and (ii) conditions of Rolle's theorem is satisfied. f (2) = 2 2 - 8 (2) + 12 = 0 f (6) = 36 - 48 + 12 = 0 ..