If ƒ(x) = x3
- 3x2
- 45x + 25 , then ƒ (x) has maximum at x =
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Given : f(x) = x³ - 3x² - 45x + 25
To Find : ƒ (x) has maximum at x =
Solution:
f(x) = x³ - 3x² - 45x + 25
=> f'(x)= 3x² - 6x - 45
f'(x) = 0
=> 3x² - 6x - 45 = 0
=> x² - 2x - 15 = 0
=> x² - 5x + 3x - 15 = 0
=> (x - 5)(x + 3) = 0
=> x= 5 or - 3
f'(x)= 3x² - 6x - 45
=> f''(x) = 6x - 6
= 6 (x - 1)
x= 5
f''(x) = 6(5 - 1) = 24 > 0 Hence minimum value
x = - 3
f''(x) = 6(-3 -1 ) = - 24 < 0 hence maximum value
ƒ (x) has maximum at x = - 3
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