absolute minimum and maximum

Question

f(x) = 2x^3+18x^2-162x+11 with -9< x< 4

This function has an absolute minimum the point ...

and an absolute maximum at the point ...

note : both parts of this answer should be entered as an ordered pair , including the parentheses, such as (5, 11 )

Answer 1

f(x) = 2x^3 +18x^2 - 162x + 11

take the derivative and set equal to zero, then solve for x

f'(x) = 6x^2 + 36x - 162 = 0

6x^2 +36x-162= 0

divide by 6

x^2+6x -27 = 0

factor

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

x=3, -9 are x coordinates of local extrema

plug them into the original cubic function to find corresponding y coordinates

y = 2(27)+18(9)-162(3)+11= 54+162-486+11=227-486=-259

and

y=2(-729)+18(81)-162(-9) +11=-1458+1458+ 1458+11= 1469

local max is (-9,1469), local min is (3,-259)

then test the end points of the interval

f(4)= 2(64)+18(16)-126(4)+11 = 128+288- 604+11= 416-593=-177>-259

(3,-259) would then be the absolute minimum

with (-9,1469) the absolute maximum in the interval

check the arithmetic, which gets a little tedious. No guarantees this is error free

but it's the basic method

another approach is use a graphing calculator online or hand held