curve sketching: x^5 - 5x^4, x^2/(x^2 + 1)

[Guest]

can you possible graph:
x^5-5x^4

x^2/(x^2+1)

so I can check my answer?
I have more questions I want to check my answer with so is there a site with a graphing calculator on it?

thanks,
take care.

 

[galactus]

Here ya' go. Also, here is a link to download the graphics software I used to create this graph. It's very nice, user-friendly, and best of all....it's free.


http://www.padowan.dk/graph/

 

[stapel]

so I can check my answer?

Maybe some time you could show your work, if you want it to be checked.

And when you post these algebra-type questions in the "Calculus" category, it might be nice if you specified the calculus stuff you're supposed to be including in your solution.

Thank you.

Eliz.

 

[soroban]

Hello, bittersweet!

Graph: \(\displaystyle y \;= \;x^5\,-\,5x^4\)

x-intercepts: \(\displaystyle \;x^5\,-\,5x^4\:=\:0\;\;\Rightarrow\;\;x^4(x\,-\,5)\:=\:0\;\;\Rightarrow\;\;x\:=\:0,\;5\)

Extrema: \(\displaystyle \,5x^4\,-\,20x^3\:=\:0\;\;\Rightarrow\;\;5x^3(x\,-\,4)\:=\:0\;\;\Rightarrow\;\;x\:=\:0,\;4\)

Using the second derivative test, we find that:
\(\displaystyle \;\;\;(0,0): \text{ maximum}\)
\(\displaystyle \;\;\;(4,-256): \text{ minimum}\)

That should be enough to graph the function . . .

              |
              |                        *
              |
      --------o-----------------------o----
          *   |  *                     5
        *     |     *                *
       *      |                     *
              |         *          *
      *       |           *      *
              |              o
     *        |           (4,-256)

Graph: \(\displaystyle \L\,y\;=\;\frac{x^2}{x^2\,+\,1}\)

The only intercept is (0,0).

First derivative: \(\displaystyle \L\,y'\;=\;\frac{(x^2\,+\,1)\cdot2x\,-\,x^2\cdot2x}{(x^2\,+\,1)^2}\;=\;\frac{2x}{(x^2\,+\,1)^2}\)
\(\displaystyle \;\;\)The only critical point is \(\displaystyle (0,0)\), a minimum.

Second derivative: \(\displaystyle \L\,y''\;=\;\frac{2(1\,-\,3x^2)}{(x^2\,+\,1)^3}\)
\(\displaystyle \;\;\)Inflection points: \(\displaystyle \,\left(\pm\frac{\sqrt{3}}{3},\,\frac{1}{4}\right)\)

Horizontal asymptote: \(\displaystyle \,y\,=\,1\)

                          |
      - - - - - - - - - - + - - - - - - - - - -
          *               |               *
              *           |           *
                 *        |        *
                   *      |      *
                    *     |     *
                      *   |   *
      --------------------o--------------------
                          |

See?
No graphing calculator could have told you all this.

 

[Guest]

"First derivative: The only critical point is (0,0), a minimum. "

I got this for the first derivative:
y'= 2x/x^2+1) -[x^2/x^2+1)^2
y'= [2x(x^2+1)-x^2]/[x^2+1)^2]
y'= [2x^3+2x-x^2]/(x^2+1)^2
0=[x(2x^2+2-x)]/[(x^2+1)^2]
x=0, then I tried to factor the (2x^2-x+2) using quad formula but got to 1+,- ( -15squareooted)/4 so you cant get imaginary numbers right..does that mean that critical number just doesn't exists when you try to solve for x but get a negative inside a squareroot?

thanks for the graphs. My first graph looked like sorobans..not really like the graphing calculator one but its the same thing, I've yet to graph y= x^2/(x^2+1) because I think I may have done the critical step part one wrong. (check above)