Question

12.5.67 Find the absolute maximum and​ minimum, if either​ exists, for the function on the indicated interval. f (x )equals 2 x cubed minus 36 x squared plus 210 x plus 1 ​(A) [negative 2 comma 9 ]​(B) [negative 2 comma 7 ]​(C) [6 comma 9 ]gFind the absolute maximum is ____at x=___

Answer 1

absolute maximum is 401 at x = 5

Explanation:

Data provided in the question:

f(x) = 2x³ - 36x² + 210x + 1 ...........(1)

To find the point of maxima or minima, differentiating the function and equating it to zero

thus,

f'(x) = (3)2x² - (2)36x + 210 + 0 = 0 ............(2)

or

6x² - 72x + 210 = 0

or

6(x² - 12x + 35) = 0

or

x² - 12x + 35 = 0

x² - 5x - 7x + 35 = 0

or

x(x - 5) - 7(x - 5) = 0

or

(x - 5)(x - 7) = 0

therefore,

the point of maxima and minima are

x - 5 = 0 and x - 7 = 0

or

x = 5 and x = 7

again differentiating (2) to find whether the points are maxima or minima

f"(x) = (2)(3)2x - (2)36

or

f"(x) = 12x - 72

at x = 5

f"(5) = 12(5) - 72

or

f"(5) = - 12

since , f"(5) is negative, the point is of maxima

similarly,

f"(7) = 12(7) - 72 = 12

since , f"(7) is positive, the point is of minima

therefore,

absolute maximum = f(5) = 2(5)³ - 36(5)² + 210(5) + 1

= 2(125) - 36(25) + 1050 + 1

= 250 - 900 + 1050 + 1

= 401

Hence,

absolute maximum is 401 at x = 5