Use a graphing utility to approximate the local maximum value and local minimum value of the function f(x) = -0.3x^3-0.6x^2+4x-6 for -6 < x < 4.

Question

The local Maximum is y = and occurs at x =

The local minimum is y = and it occurs at x =

Answer 1

-0.3x^3 - 0.6x^2 + 4x -6, and -6 < x < 4

take the derivative and set = 0

-.9x^2 - 1.2x + 4 = 0, solve for x, use the quadratic formula

1.2/-1.8 + or - (1.8)sqr(1.44 + 4(.9)(4)) = - 2/3 + or - 3.98/1.8 = 1.54, -2.88

those are the two turning points in the cubic equation's graph

f(x)=-about f(-2.88) is the local minimum =- .3(2.88)^3 - .6(2.88)^2 + 4(-2.88) - 6 = -15.33 = y

x = about 1.54. f(1.54) is the local maximum = -.3(1.54)^3 -.6(1.54)^2 +4(1.54) -6 = -1.1 - 1.42 +6.16 - 6 = -1.92 = y

local min: y=-15.33, when x = -2.88

local max: y=-1.92 when x = 1.54

check either side of the x values with simple integers

f(-3) = -.3(-3)^3 -.6(-3)^2 +4(-3)-6 = 8.1 -5.4 -12-6 = -15.3 < f(-2.88)=-15.33

f(-2) = -.3(-2)^3 - .6(-2)^2 +4(-2) - 6 = 2.4 -2.4 - 8 -6 = -14 < f(-2.88) = 15.33

f(2) = -.3(2)^3 - .6(2)^2 + 4(2) - 6 = -2.4 -2.4 + 8 -6 = -2.4 < f(1.54) = -1.92

f(1) = -.3 - .6 + 4 -6 = -2.9 < f(1.54) = -1.92

Use a graphing calculator, but whatever it shows, check it against the above

y = -15.33 when x = -2.88, y= -1.92 when x = 1.54