Respuesta :
Since the function is continuous between x = 0 and x = 44 then Rolle's theorem applies here.
Differentiating
y' = x * 2(x - 44) + (x - 44)^2
y' = 3x^2 - 176x + 1936 = 0 (at a turning point).
solving we get x = 44 , 14.67
y" = 6x - 176 which is negative for x = 14.67 so this gives a maximum value for f(x)
This maximum is at the point (14.67, 12,619.85)
There is a minimum at ( 44,0)
These are the required points
Rolle's theorem function f is continuous upon that closed interval [a, b] and distinguishable upon that open interval (a, b) as such f(a) = f(b), therefore f′(x) = 0 some of x with [tex]\bold{ a\leq x\leq b}[/tex], and further calculation can be defined as follows:
Given:
[tex]\bold{f(x)=x(x-44)^2 \ \ \ \ \ \ \ \ \ \ [0,44]}[/tex]
To find:
Determine by Rolle's theorem=?
Solution:
By Rolle's theorem will have [tex]\bold{f'(c) = 0}[/tex]
[tex]\bold{f'(x)= 2x(x-44)+(x-44)^2=0 \ \ \ \ \ \ \ \ \ \ [0,44]}\\\\[/tex]
[tex]\bold{(x-44) (2x+(x-44))=0 }\\\\\bold{(2x+(x-44))=0 }\\\\\bold{2x+x-44=0}\\\\ \bold{3x-44=0}\\\\\bold{x=\frac{44}{3}}\\\\[/tex]
Therefore, [tex]\bold{f'(\frac{44}{3})=0}[/tex]
Learn more:
brainly.com/question/2292493
