# The graph of a function f is linked below. Sketch the antiderivative F(x) of f defined on the interval [0,3] and such that F(0)= 0. The graph of a function f is sketched below. Sketch the...

The graph of a function f is linked below. Sketch the antiderivative F(x) of f defined on the interval [0,3] and such that F(0)= 0.

The graph of a function f is sketched below. Sketch the antiderivative F(x) of f defined on the interval [0,3] and such that F(0)= 0. *(Hint: **What does *F(x)* represent on the graph of *f(t)?*)*

the links to the two graphs of function f are:

http://s19.postimage.org/oox9wvk8j/calc_hw_23_9_graph_1.png

http://s19.postimage.org/g97pfdhdf/calc_hw_23_9_graph_2.png

### 1 Answer | Add Yours

Some general things to be aware of:

Whenever f(x)=0, that is a place where F "flattens out"

Whenever f(x) is positive, then F(x) is increasing, and whenever f(x) is negative, F(x) is decreasing, so given your graph we want F to be:

DECREASING from x=-2 to x =0

FLATTENING OUT at x=0

INCREASING from 0 to 1/6

FLATTENING OUT at 1/6

INCREASING from 1/6 up to 1/2 (I'm guessing at the 1/2)

FLATTENING OUT at x=1/2

INCREASING from 1/2 to 3/2

FLATTENING OUT at 3/2

and INCREASING from 3/2 and on

also, we want F(0)=0 (because we are told that from the problem)

even better though, your graph is made out of linear pieces, which we can integrate (to get quadratics)

I am assuming the following, although it is a little difficult to tell from the picture, that the pieces connect the following pairs of points:

(-2,-2) to (1/9,1/9)

(1/9,1/9) to (1/6,0)

(1/6,0) to (1/3,1/3)

(1/3,1/3) to (1/2,0)

(1/2,0) to (1,1)

(1,1) to (3/2,0)

(3/2,0) to (3,3)

In which case, the function of your graph is given by:

`y=`

`x " if " -2 <= x < (1)/(9)`

`-2(x-(1)/(6))=-2x+(1)/(3) " if " (1)/(9) <= x < (1)/(6)`

`2(x-(1)/(6))=2x-(1)/(3) " if " (1)/(6) <= x < (1)/(3)`

`-2(x-(1)/(2))=-2x+1 " if " (1)/(3) <= x < (1)/(2)`

`2(x-(1)/(2))=2x-1 " if " (1)/(2) <= x < 1`

`-2(x-(3)/(2) ) = -2x + 3 " if " 1 <= x < (3)/(2)`

`2(x-(3)/(2) ) = 2x - 3 " if " (3)/(2) <= x < 3`

We can integrate:

`F(x)=`

`(1)/(2)x^2 +C_1 " if " -2 <= x < (1)/(9)`

`-x^2+(1)/(3)x+C_2 " if " (1)/(9) <= x < (1)/(6)`

`x^2-(1)/(3)x+C_3 " if " (1)/(6) <= x < (1)/(3)`

`-x^2+x+C_4 " if " (1)/(3) <= x < (1)/(2)`

`x^2-x+C_5 " if " (1)/(2) <= x < 1`

`-x^2+3x+C_6 " if " 1 <= x < (3)/(2)`

`x^2-3x+C_7 " if " (3)/(2) <= x < 3`

We want F(x) to go through (0,0), so we must have C1 = 0

We can connect each of the pieces to this first piece to get our other C's:

`F(x)=(1)/(2)x^2 +0 " if " -2 <= x < (1)/(9)`

so `F( (1)/(9) ) = (1)/(162)`

`F(x) = -x^2+(1)/(3)x+C_2 " if " (1)/(9) <= x < (1)/(6)`

so `C_2 = -(1)/(54)`

`F(x) = -x^2+(1)/(3)x-(1)/(54) " if " (1)/(9) <= x < (1)/(6)`

so `F( (1)/(6) ) = (1)/(108)`

`x^2-(1)/(3)x+C_3 " if " (1)/(6) <= x < (1)/(3)`

so `C_3 = (1)/(27)`

`F(x)=x^2-(1)/(3)x+(1)/(27) " if " (1)/(6) <= x < (1)/(3)`

so `F( (1)/(3) ) = (1)/(27)`

`-x^2+x+C_4 " if " (1)/(3) <= x < (1)/(2)`

so `C_4 = (-5)/(27)`

`F(x)=-x^2+x-(5)/(27) " if " (1)/(3) <= x < (1)/(2)`

so `F( (1)/(2) ) = (7)/(108)`

`F(x) = x^2-x+C_5 " if " (1)/(2) <= x < 1`

so `C_5 = (17)/(54)`

`F(x) = x^2-x+(17/54) " if " (1)/(2) <= x < 1`

(You can continue on in this way to get the other unknown constants.)

And the graph (zoomed in to see the important features) looks like: