the
find the maximum and minimum
function x² + 2x² - 4x +8
Answers
Answer :
• Maximum value of f(x) = 16
• Minimum value of f(x) = 6 14⁄27
Solution :
[ Using 1st derivative test ]
• Given function : x³ + 2x² - 4x + 8
• To find : Minimum and maximum values of the given function .
Let the given function be f(x) .
Thus ,
f(x) = x³ + 2x² - 4x + 8
Now ,
The 1st order derivative of f(x) will be ;
f'(x) = 3x² + 4x - 4
We know that ,
At points of minima or maxima of a function , its first derivative is zero .
Thus ,
=> f'(x) = 0
=> 3x² + 4x - 4 = 0
=> 3x² + 6x - 2x - 4 = 0
=> 3x(x + 2) - 2(x + 2) = 0
=> (x + 2)(3x - 2) = 0
=> x = -2 , ⅔
• At x = -2 , the functional value will be ;
=> f(-2) = (-2)³ + 2•(-2)² - 4•(-2) + 8
=> f(-2) = -8 + 8 + 8 + 8
=> f(-2) = 16
• At x = 2/3 , the functional value will be ;
=> f(⅔) = (⅔)³ + 2•(⅔)² - 4•(⅔) + 8
=> f(⅔) = 8/27 + 8/9 - 8/3 + 8
=> f(⅔) = (8 + 8•3 - 8•9 + 8•27) / 27
=> f(⅔) = (8 + 24 - 72 + 216) / 27
=> f(⅔) = 176/27
=> f(⅔) = 6 14⁄27
Clearly , f(-2) > f(⅔)
Hence ,
• f(-2) = 16 is maximum value of f(x) and x = -2 is the point of maxima .
• f(⅔) = 6 14⁄27 is the minimum value of f(x) and x = ⅔ is the point of minima .