Pl. note correction in absolute minimum.

**Function has absolute minimum=-76 at x=-3**

Given: `f(x)=x^3-6x^2+5,[-3,5].`

Find the critical value(s) by setting the first derivative equal to zero and solving for the x value(s).

`f'(x)=3x^2-12x=0`

`3x(x-4)=0`

`x=0,x=4`

The critical numbers are x=0 and x=4 . Plug in the critical numbers and the endpoints of the interval [-3, 5] into the original f(x) function.

f(-3)=-76

f(0)=5

(The entire section contains 3 answers and 233 words.)

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