Question

(a) Find the critical numbers of the function f(x) = x(x - 1)? (smallest value) X= X (largest value) (b) What does the Second Derivative Test tell you about the behavior off at these critical numbers? At x = the function has --Select... (c) What does the First Derivative Test tell you? (Enter your answers from smallest to largest x value.) At x = the function has --Select- At x = the function has --Select- At x = the function has ---Select-

Answer 1

8 x We have: f(x) (SC-1) 7 To find the critical values of f(x), we solve f'(x) = 0 a.) scº ( 7 (x416) + 8x* (x417 = 0 x? (x4) 6 [7x + 8(x64)] = 0 sc? (scyla [ 15x -8] = 0 -) Sc = 0, 1, are the critical points =) =) -) 8 15 0 JC = 10 (smallest) o ans. 8/15 x 1 (largest) ( b ) = f" (XC) X7 (x4) (15) + x + (15x-8) (6) (x4)5. 7x6 (x4) 6 (15X-8) x6 (64)5 [ x2(x+) (15) + x(x4) (15) + x 16) (153-8) + +(x4) (15X-8) (No conclusion) ciis 1" (it) = (13) (i) (p) ( ) (15) + 6(3) (0) + 7(is (0) Now oil faco) Be + 6 value at 8 15 f(x) has a minimum. CS Scanned with CamScanner

At x=1, f"W = 0 (No conclusion) Hence At = 8 by 1-6 derivative test : the function has a minimum) 15 aus ans We have : (C. f (-1) = ()7 (-2) 6 (-23) = ③ value. > 06 OK o f' 10) f' (05) = (05j7 6.5)6 (7.5 -8) = o value. et BE D 6 ' (=.54) (3) (i) (O) (075)} (-0.25)6 (11.25-8) = o value f' (:75) u > 1. (1) ľ (lot) = (1+r)? (01)6 (16.5-8)= value o f'(x) =) 0 8 1 CS Scanned with CamScanner

Now f'(sc) changes sign from 0 to o at xc=0 Ax (x=0 a [maxima , function has 15 y of'(xc) changes sign from o to 4 at se = 8 x = 8), function has a Minimal =) At x= 1, l'(x) doesn't change sign function has a saddle point). At c=1 CS Scanned with CamScanner