Raymond B. answered • 01/25/22
Math, microeconomics or criminal justice
x^3/2 - x^2/2 -3x + 2
f(0) = f(1) = 2
f(1) = -1
an x intercept is in the interval 0<x<1, where f(x) changes sign. there is between 1 and 3 real x-intercepts in that interval. if just one, then the other 2 x-intercepts are imaginary. imaginary zeros come in conjugate pairs.
as x gets larger, f(x) gets increasingly positive & approaches infinity
as x gets smaller, f(x) gets more & more negative and approaches negative infinity
as a cubic polynomial there are 3 zeros, although 2 may be imaginary.
any local extrema are where the derivative or slope = 0
f'(x) = 3x^2/2 - x - 3 = 0
x = 1/3 + or - (1/3)sqr(1+18) = about -1.1 & 1.8
local max is about (-1.1, f(-1.1)) = (-1.1, 4 )
local min is about (1.8, f(2.3)) = (1.8, -0.3)
f(-1.1) = -.7 -.6 + 3.3 + 2 =4
f(1.8) = 2.9 -1.6 - 3.6 +2 = -0.3
one zero is between the local max and min, and in the interval 0<x<1 it's about 0.
(1/2)(.6)^3 - (1/2)(.6)^2 - 3(.6) + 2 = .1 -.2 - 1.8+ 2 = close to zero
divide the cubic by x-.6 to get a quadratic, set it equal to zero, solve for x to get an estimate of the other two zeros about -2 and 2.8
f(2) = -2
f(3) = 2
one zero is in the interval 2<x<3 where f(x) changes sign
f(-2) = 2
f(-3) = -7
another zero is in the interval -3<x<-2 where f(x) changes sign
If you want to solve to the nearest hundredth, the calculations get very tedious. It makes you wonder if the problem was copied correctly. the cubic polynomial doesn't have easily calculated zeros. But given as written, whatever you come up with should be in the ranges above. 3 real zeros, one negative, two positive. one between the turning points, the local max and min, and one to the right of the local min, the other to the left of the local max
