Calculus Question

Question

Find the absolute minimum and absolute maximum values of f on the given interval.

f(x) = (x² − 1)³,    [−1, 4]

absolute minimum
absolute maximum

Answer 1

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

take the derivative and set = 0, solve for x

f'(x) = 6x^5 -12x^3 + 6x = 6x(x^4 - 2x^2 + 1) = 6x(x^2-1)^2 = 6x(x+1)^2(x-1)^2 = 0

set each factor = 0

x= 0, -1, 1 are the 3 zeros, the 1st with multiplicity 1,

the last 2 with multiplicity 2 each, where the graph is tangent to the x axis but does not cross it. at x=0 the graph crosses the x axis

for a 6th degree polynomial there are 5 potential zeros or roots or x intercepts, 1 less than the degree of the polynomial.

check the endpoints and the local extrema

f(0 ) = 1

f(1) = 0

f(-1) = (-1)^3 = -8 = absolute minimum on the interval [-1, 4]

f(4) = (4-1)^3 = (3)^3 = 27 = absolute maximum on the interval [-1,4]

absolute minimum is (-1, -8) in the given interval

absolute maximum is (4, 27) in the given interval

relative or local maximum is (0,1)

relative or local minimum is (1,0)