can anybody help withthis question on absolute extrema??

Question

a) Listed below are steps in the procedure to find absolute extrema for a function f that is continuous on a closed interval [a,b]. Put them in the right order.

1 2 3 Find the values of f at the critical numbers of f in (a,b).
1 2 3 Find the values of f at the endpoints of the interval.
1 2 3 The absolute maximum of f on [a,b] is the greatest of the values of f at the endpoints a and b and at the critical values on (a,b).
The absolute minimum of f on [a,b] is the least of the values of f at the endpoints a and b and at the critical values on (a,b).

(b) Find absolute extrema for f(x)=3 x3 −18 x2+27 x−5 over the closed interval [2,4].
Solution: The absolute maximum is f(?)= ?.
The absolute minimum is f(?)= .?

Answer 1

(b) f(x) = 3x^3 - 18x^2 + 27x - 5

First, we find local extrema by finding the first derivative and then making it 0.

f'(x) = 9x^2 - 36x + 27
f'(x) = 9(x^2 - 4x + 3)
f'(x) = 9(x - 3)(x - 1)

Now, make f'(x) = 0.

0 = 9(x - 3)(x - 1)

Which means our critical values are x = {1, 3}

Note that [2, 4] is our interval, and 1 does not fall within the interval, so we discard 1.

Now, we calculate our critical point at our function.
f(3) = 3(3)^3 - 18(3)^2 + 27(3) - 5
f(3) = 3*27 - 18*9 + 27*3 - 5
f(3) = 6*27 - 18*9 - 5
f(3) = 162 - 162 - 5 = -5

And we also test our endpoints.

f(2) = 3(2)^3 - 18(2)^2 + 27(2) - 5
f(2) = 24 - 72 + 54 - 5
f(2) = 1

f(4) = 3(4)^3 - 18(4)^2 + 27(4) - 5
f(4) = 192 - 288 + 108 - 5
f(4) = 7

Now, compare f(3) = -5, f(2) = 1, and f(4) = 7

Absolute minimum at (3, -5).
Absolute maximum at (4, 7).

Answer 2

The steps are already in a workable order as written, although you might want to reverse #s 1 & 2.
f(x) = 3x^3 − 18x^2 + 27x − 5
f(2) = 24 - 72 + 54 - 5 = 1
f(4) = 192 - 288 + 108 - 5 = 7
9x^2 − 36x + 27 = 0
x^2 − 4x + 3 = 0
(x - 1)(x - 3) = 0
x = 1, 4
81 − 162 + 81 − 5 = - 5
on the interval[2,4]
absolute maximum is (4,7)
absolute minimum is (3,- 5)
(x = 1 is outside the interval)