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3 years ago

Find the absolute maximum and absolute minimum values of f on the given interval. (If an answer does not exist, enter DNE.)

Mathematics

1 answer:

vodka [1.7K]3 years ago

7 0

Answer:

absolute max is 120 and absolute min is -8

Step-by-step explanation:

Find critical numbers

f'(x) = 3x^2 - 12x + 9 = 0

= 3(x^2 - 4x + 3) = 0

3(x-3)(x-1) = 0

(x-3) = 0 or (x-1)=0

x = 1,3

Test them!

x<1 Sign of f' on this interval is positive

1<x<3 Sign of f' on this interval is negative

x>3 Sign of f' on this interval is positive

f(x) changes from positive to negative at x = 1 which means there is a relative maximum here.

f(x) changes from negative to positive at x = 3 which means there is a relative minimum here.

Test the endpoints to find the absolute max and min.

f(-1) = -8

f(1) = 12

f(3) = 8

f(7) = 120

The absolute maximum value of f is 120 and the absolute minimum value of f is -8.

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2 years ago

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Answer:

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Step-by-step explanation:

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7 0

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Answer:

Step-by-step explanation:

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