(x) = x4 − 2x3 + x + 1, [−1, 3]

Question

Find the absolute extrema of the function on the given interval. (Round your answers to three decimal places. If an answer does not exist, enter DNE.)

minima (x,y)=? (smaller x-value)

(x,y)=? (larger x-value)

maximum (x,y)=?

Answer 1

x^4 - 2x^3 + x + 1 in the interval [-1, 3]

f(-1) = (-1)^4 -2(-1)^3 -1 +1 = 1+2 -1+1 = 3

f(3) = (3)^4 -2(3)^3 +3 + 1= 81 -54 +4= 31

(x,y) = (3,31) looks like the likely absolute maximum in the given interval

(x,y) = (-1,3) looks like the likely absolute minimum in the given interval, Unless:

the 4th degree equation has potentially 3 turning points and 1, 2 or 3 of them may have a local max or min, smaller or larger than the end points of the interval.

f(x) = x^4 -2x^3 + x + 1

take the derivative

f'(x) = 4x^3 -6x^2 +1

set it equal to zero and solve for x to find a turning point

f'(1/2) = 4(1/8) - 6(1/4) + 1 = 1/2 -3/2 +1 = 0

f(1/2) = 1/16 -2/8 +1/2 + 1 = 5/4+1/16 =21/16 which is <31 and <3,

divide x-1/2 into f'(x) to get 4x^2 -4x -2

set it equal to zero

4x^2 -4x -2 = 0

divide by 2

2x^2 -2x -1 = 0

solve for x

x - about 1.367 and -0.367

f(1.367) = about 0.706

f(-0.37) = about 1.65

(1.37, 0.706) is a relative minimum and the absolute minimum

it helps to try to sketch a rough graph, & plot more points

plot some more points in the interval -1<x <3

x f(x)

-1 3

-.37 1.65 a turning point

-1/2 13/16

0 1

1/2 27/16 a turning point, and relative maximum

1 1

1.37 0.706 a turning point, relative & absolute minimum

3/2 13/16

2 3

3 31 absolute maximum

absolute maximum is (3, 31)

absolute minimum is about (1.367, 0.75)

for the interval [-1, 3}

for the interval -inf < x < +inf, [negative infinity to positive infinity] there is no absolute maximum, f(x) approaches positive infinity but never reaches it.