Let f(x) = x^4 + 0.2x^3 - 5.8x^2 - x + 4, x∈R. (a) find the solutions of f(x) > 0. (b) for the curve f(x), (i) find the coordinates of both local minimum points. (ii) find th…

Question

Let
`f(x) = x^4 + 0.2x^3 - 5.8x^2 - x + 4`, x∈R.

(a) find the solutions of f(x) > 0.
(b) for the curve f(x),
(i) find the coordinates of both local minimum points.
(ii) find the x-coordinates of the points of inflection.

Answer 1

Hi,

Explanation:

`x^4+0.2x^3-5.8x^2-x+4\\=x^4+x^3-0.8x^3-0.8x^2-x^2 -x -4x^2+4\\=x^3(x+1)-0.8x^2(x+1)-x(x+1)-4(x^2-1)\\=(x+1)(x^3-0.8x^2-x-4(x-1))\\=(x+1)(x^2(x-0.8)-5(x-0.8))\\=(x+1)(x-0.8)(x^2-5)\\\\=\boxed{(x+1)(x-0.8)(x-√(5) )(x+√(5) )}\\`

a)

b)

`sol=\{-√(5),-1,0.8,√(5) \}`

c)

`f'(x)=4x^3+0.6x^2-11.6x-1=0\\\\x_1 = 1.672875364844 \\x_2 = -1.73683174462 \\x_3 = -0.086043620224\\\\\\`

minimum are -3.70 for x=-1.74, and -5.14for x=1.67

points of inflexion are

`f''(x)=12x^2+1.2x-11.6=0\\x=-0,93\ ,\ y=0.49\\x=0.93\ ,\ y=-1.05\\`