# The graph of f(x)= (1/2)(1-x)((x^2)-x-2) is shown at the following link: http://s19.postimage.org/5nxj1rj8z/graph_for_calc_hw_26_23.png .The graph of f(x)= (1/2)(1-x)((x^2)-x-2) is shown at the...

The graph of f(x)= (1/2)(1-x)((x^2)-x-2) is shown at the following link: http://s19.postimage.org/5nxj1rj8z/graph_for_calc_hw_26_23.png .

The graph of f(x)= (1/2)(1-x)((x^2)-x-2) is shown at the following link: http://s19.postimage.org/5nxj1rj8z/graph_for_calc_hw_26_23.png .

(a) Find the X and Y intercepts.

(b) Determine the intervals on which f is decreasing and intervals on which it is increasing.

(c) Find local minima and local maxima of f(x).

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You may find x intercepts solving the equation `(1/2)(1-x)(x^2-x-2) = 0` such that:

`(1/2)(1-x)(x^2-x-2) = 0 =gt 1-x = 0 =gt `

`x = 1

`` `` x^2 - x - 2 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (1+-sqrt(1 + 8))/2 =gt x_(1,2) = (1+-3)/2 `

`x_1 = 2 ; x_2 = -1`

**Hence, the graph of function intercepts x axis at (-1,0) ; (1,0) and (2,0).**

You may find y intercepts putting x = 0 such that:

`f(0)= (1/2)(1-0)(0-0-2) =gt f(0) = -1`

**Hence, the graph of function intercepts y axis at (0,-1).**

You need to determine the local maxima or minima of the function, hnece, you should solve the equation f'(x) = 0 such that:

`f'(x) = (-1/2)(x^2-x-2) + (1/2)(1-x)(2x-1) `

`f'(x) =(1/2)(-x^2 + x + 2 + 2x - 1 - 2x^2 + x) `

`f'(x) =(1/2)(-3x^2 + 4x + 1)`

`f'(x) = 0 =gt -3x^2 + 4x + 1 = 0 =gt 3x^2 - 4x - 1 = 0 `

`x_(1,2) = (4+-sqrt(16 + 12))/6 `

`x_(1,2) = (4+-2sqrt7)/6`

**Hence, the function reaches its maximum at `x = (2-sqrt7)/3` and it reaches its minimum at `x = (2+sqrt7)/3` .**

**Notice that the function increases over `(-oo;(2-sqrt7)/3)U((2+sqrt7)/3;oo` ) and it decreases over `((2-sqrt7)/3 ; (2-+sqrt7)/3).` **