f(x) = x3 - 4x2 + 5x - 2
f'(x) = 3x2 - 8x + 5
put f'(x) = 0
3x2 - 8x + 5 = 0
3x2 - 3x - 5x + 5 = 0
3x(x -1) - 5(x - 1) = 0
(3x - 5)(x -1) = 0
3x - 5 = 0 or x - 1 = 0
x = 5/3 or x = 1
critical numbers are
x = 5/3
x = 1
Second derivative test
f'(x) = 3x2 - 8x + 5
f''(x) = 6x - 8
at x = 1
f''(1) = 6(1) - 8 = 6 - 8 = - 2 < 0
at x = 5/3
f''(5/3) = 6(5/3) - 8 = 10 - 8 = 2 > 0
Thus x = 1 is point of local maximum or at x =1 f(x) is local maximum and x = 5/3 is point of local minimum or f(x) is local minimum at x = 5/3
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