The graph of f(x)= (1/2)(1-x)((x^2)-x-2)  (d) Determine concavity intervals for f. Determine inflection points (e) Determine  asymptotes if any? The graph of f(x)= (1/2)(1-x)((x^2)-x-2)  is...

The graph of f(x)= (1/2)(1-x)((x^2)-x-2)  (d) Determine concavity intervals for f. Determine inflection points

(e) Determine  asymptotes if any?

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 .

(d) Determine concavity intervals for f. Determine inflection points on the graph y=f(x)

(e) Determine (vertical and horizontal) asymptotes if any?

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

d) You should remember that the second order derivative tells you where the graph of function is concave up or concave down.

You need to evaluate first the first order derivative using the product rule such that:

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

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

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

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

You need to differentiate the first order derivative to find the second order derivative such that:

`f''(x) = (1/2)(-6x + 4) => f''(x) = -3x + 2`

You may find the inflection point solving the equation `f''(x) = 0`  such that:

`-3x + 2 = 0 =gt -3x = -2 =gt x = 2/3`

Hence, the function changes its concavity at `x = 2/3` .

Notice that the second order derivative is positive over `(-oo,2/3), ` hence the graph is concave up if `x in (-oo,2/3)` . The second order derivative is negative  over `(2/3 , oo), ` hence the graph is concave down if `x in (2/3,oo).`

e) Since the domain of the function has no restriction, hence, there is no vertical asymptotes.

You need to evaluate the horizontal asymptotes, if any, hence, you should evaluate the following limits such that:

`lim_(x->-oo) f(x) = lim_(x->-oo) (1/2)(1-x)((x^2)-x-2) `

`lim_(x->-oo)(1/2)(1-x)((x^2)-x-2) = (1/2)lim_(x->-oo) (-x^3 + x^2 + 2x + x^2 - x - 2)`

`lim_(x->-oo) f(x) = (1/2)lim_(x->-oo) (-x^3 + 2x^2 + x - 2)`

`lim_(x->-oo) f(x) = (1/2)(oo+oo+oo-2) = oo`

Hence, evaluating the horizontal and vertical asymptotes yields that there are no horizontal and vertical asymptotes. 

We’ve answered 318,917 questions. We can answer yours, too.

Ask a question