Answer:
The coordinate of the relative maximum is x=4.
Step-by-step explanation:
Given that the derivative of the function [tex]f(x)[/tex] is [tex]f'(x) = (x-4)(4-2x)[/tex], the maxima and minima or the critical points can be found where [tex]f'(x)=0,[/tex] that is:
[tex]f'(x) = (x-4)(4-2x)=0.[/tex]
The solutions to this equation are [tex]x=4[/tex] and [tex]x=2.[/tex]
Now, if the second derivative [tex]f''(x)[/tex] for a function [tex]f(x)[/tex] is negative at a critical point, then the critical point is the relative maximum.
Therefore we want to see at which of the two critical points is [tex]f''(x)[/tex] negative. The second derivative is:
[tex]\frac{\mathrm{d f'(x)}}{\mathrm{d}x}=f''(x)=-4(x-3).[/tex]
Now [tex]f''(4)[/tex] is [tex]-4[/tex], and [tex]f''(2)[/tex] is [tex]+4[/tex], therefore we deduce that the relative maxium is located at [tex]x=4[/tex], because there the second derivative [tex]f''(x)[/tex] is negative.
