Question

Please help!

Verify that Rolle's Theorem can be applied to the function f (x) = 2,3 - 1022 +312 – 30 on the interval (2,5). Then find all values of c in the interval such that f'(c) = 0. Enter the exact answers in increasing order. To enter Vā, type sqrt(a).

Answer 1

f(x)= x² - 103c² +312-30 C2,5] f(x) is Polynomal function hence it is Continnous and differentiable every where So, f(x) is Continuous on [25] f (sc) is differentiable an (2,5) f(2)= 22 - 10x2² + 3142 -30 = 0 f(s) = 5 - 10 x 5² +31 x5 -30 = 880 - 280 on DS(2) = f(5) .: All the three hypothesis of Rolle's theorem is Varified i Rolle's theorem can be an applied to foi [2,5 Now, f(x) = 2(²-10x²+ 310-30 f'(x) = 3x² – 20x +31 f'(c) =36² - 200 tal f'(c) 32² 200+31=0 - (-20) + [(-20) » _ 4x3 x3) 2x3 400-372 6

8وی + و c= مه دو + نوے - G 10 + 3 C = | 0 + 4 3 | 0 - 7 7 3 But 5 ر2] ع And, 2_ 10 د + 5 3 دریا - 2 ی 17 3 ور - ح) - و حال 3 3