Answer:
The absolute minimum value of the function over the interval [0,7] is -18.
Step-by-step explanation:
The given function is
[tex]f(x)=x^2-6x-9[/tex]
Differentiate f(x) with respect to x.
[tex]f'(x)=2x-6[/tex]
Equate f'(x)=0 to find the critical points.
[tex]2x-6=0[/tex]
[tex]2x=6[/tex]
[tex]x=3[/tex]
The critical point is x=3.
Differentiate f'(x) with respect to x.
[tex]f''(x)=2[/tex]
Since f''(x)>0 for all values of x, therefore the critical point is the point of minima and the function has no absolute maximum value.
3 ∈ [0,7]
Substitute x=3 in the given function to find the absolute minimum value.
[tex]f(3)=(3)^2-6(3)-9[/tex]
[tex]f(3)=9-18-9[/tex]
[tex]f(3)=-18[/tex]
Therefore the absolute minimum value of the function over the interval [0,7] is -18.
