Find the absolute maximum value and the absolute minimum value, if any, of the function.

Question

F(x)=(1/2)x⁴-(2/3)x³-2x²+3 on [-4,3]

maximun=

minimun=

Answer 1

the graph is likely similar to a "W" shape with the local maximum in the middle

local minimums at the two points at the bottom

absolute maximum is the largest of the end points of the interval

absolute minimum is the smaller of the two local minimums

F(x) = (1/2)x4 - (2/3)x^3 -2x +3

has a maximum two positive solutions and no negative solutions by Descartes' method, other solutions are imaginary

4th degree functions have 3 potential turning points, each with zero slope for a tangent line

unless there's an inflection point, then just one turning point which would be the minimum and no local max

F'(x) = 2x^3 - 2x^2 - 2 = 0

x^3 -x^2 -1 = 0

there are 3 potential real solutions. one would the local max, the other two local minimums

but there seems to be no local max and just one local minimum, given F(x) with an inflection point

x= 1.46557 is the only real solution, using an online cubic equation calculator, the other two solutions are imaginary.

F(-4) =181 2/3

F(3) = 60

181 2/3 is the absolute maximum

F(0) = 3

F(1) = 5/6

F(-1) = 6 1/6

F(2) = 1 2/3

F(1.466) = about 0

F'(1)=-2

F'(2) =6

somewhere between 1 and 2 is a turning point, since the sign changes from x=1 to x=2

F'(1.466) = .002= about 0

F'(1.46557) = 0