Question

Find the absolute maximum and minimum values of the function over the indicated​ interval, and indicate the​ x-values at which they occur. f left parenthesis x right parenthesisequalsx squared minus 6 x minus 9​; left bracket 0 comma 7 right bracket

Answer 1

Answer:

The absolute minimum value of the function over the​ interval [0,7] is -18.

Step-by-step explanation:

The given function is

$f(x)=x^2-6x-9$

Differentiate f(x) with respect to x.

$f'(x)=2x-6$

Equate f'(x)=0  to find the critical points.

$2x-6=0$

$2x=6$

$x=3$

The critical point is x=3.

Differentiate f'(x) with respect to x.

$f''(x)=2$

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.

$f(3)=(3)^2-6(3)-9$

$f(3)=9-18-9$

$f(3)=-18$

Therefore the absolute minimum value of the function over the​ interval [0,7] is -18.