# Jessica

posted by .

Find absolute maximum and minimum for f(x)=(3x^3)−(18x^2)+27x−10 over the closed interval [2,4].

Thank you so much!!

• Jessica -

f = 3x^3−18x^2+27x−10
f' = 9x^2 - 36x + 27 = 9(x^2-4x+3) = 9(x-1)(x-3)

so, f has a relative min/max at x=3
From what you know about the general shape of cubics, it should be clear that (3,-10) is a relative min.

since
f(2) = -4
f(4) = 2,

f(3) = -10 is the absolute min within the interval
f(4) = 2 is the absolute max in the interval

## Similar Questions

1. ### Calculus (pleas help!!!)

Find the absolute maximum and absolute minimum values of the function f(x)=(x−2)(x−5)^3+11 on each of the indicated intervals. Enter -1000 for any absolute extrema that does not exist. (A) Interval = [1,4] Absolute maximum …
2. ### Calculus (pleas help!!!)

Find the absolute maximum and absolute minimum values of the function f(x)=(x−2)(x−5)^3+11 on each of the indicated intervals. Enter -1000 for any absolute extrema that does not exist. (A) Interval = [1,4] Absolute maximum …
3. ### math calculus

Find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results. (Round your answers to two decimal places. If an answer does not exist, enter DNE.) g(x) = (x^2 − 4)^2/3, [−5, …
4. ### calculus

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = e−x − e−7x, [0, 1]

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 2x^3 − 3x^2 − 72x + 7 , [−4, 5]

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = (t square root of (64 − t^2)), [−1, 8]