Answer:
3
Step-by-step explanation:
To find the relative maximum of a function the double derivative should be less than zero at the stationary point. For that we need to find the stationary value/values of the function so here goes,
[tex]f(x)=-x^3+6x^2-9x-1\\f'(x)=-3x^2+12x-9\\\\For\ stationary\ point\ f'(x)\ equals\ zero\\\\0=-3x^2+12x-9\\3x^2-12x+9=0\\x^2-4x+3=0\\\\x^2-3x-x+3=0\\x(x-3)-1(x-3)=0\\(x-3)(x-1)=0\\x=3\ ,\ x=1[/tex]
Now since we have two values of x at which the function is stationary we need to test them both for that we need to take the double derivative so here goes,
[tex]f'(x)=-3x^2+12x-9\\f''(x)=-6x+12\\now\ put\ x = 1\\f''(1)=-6(1)+12\\f''(1)=-6+12=6\\since\ f''(1)>0[/tex]
The point x = 1 is a relative minimum
But we need the relative maximum so lets try out x = 3
[tex]f''(x)=-6x+12\\f''(3)=-6(3)+12\\f''(3)=-18+12=-6[/tex]
[tex]since\ f''(3)<0[/tex]
The point x = 3 is the relative maximum because for relative maximum the double derivative should be negative or less than 0 in this case its true so the last option is your answer
