Answer:
Given that,
The function is,
[tex]f(x)=-x^2+2x-4[/tex]
To find the minimum or maximum value.
As a 1st step, we need to find the derivative of a function f(x).
Derivative of a function f'(x) is,
[tex]f^{\prime}(x)=-2x+2[/tex]
Then, consider f'(x)=0.
we get,
[tex]-2x+2=0[/tex][tex]-2x=-2[/tex][tex]x=1[/tex]
x=1 is the extremum point.
To find whether the value x=1 is minimum or maximum.
we have that, To find x=a is minimum or maximum value,
we use, if x=a-c, where f'(a-c) is positive (left point), and x=a+c, where f'(a+c) is negative (right point), then the value x=a is maximum value.
if x=a-c, where f'(a-c) is negative, and x=a+c, where f'(a+c) is positive, then the value x=a is minimum value.
where c is any positive small integer.
we consider,
x=0 (left point), Substitute in f'(x), we get
[tex]f^{\prime}(x)=-2(0)+2[/tex][tex]f^{\prime}(x)=2[/tex][tex]f^{\prime}(x)=2\text{ \lparen positive\rparen}[/tex]
Consider x=2 (right point), Substitute in f'(x), we get
[tex]f^{\prime}(x)=-2(2)+2[/tex][tex]f^{\prime}(x)=-4+2[/tex][tex]f^{\prime}(x)=-2\text{ \lparen negative\rparen}[/tex]
Hence x=1 is the maximum value.
The function has a maximum value.
we get that, when x=1, f(x) is,
[tex]f(1)=-1^2+2(1)-4[/tex][tex]f(1)=-5+2[/tex][tex]f(1)=-3[/tex]
The function's maximum value is -3.
The maximum value occurs at x=1.
Answer is:
1) The function has a maximum value.
2) The function's maximum value is -3.
3) The maximum value occurs at x=1.
