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

G

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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? :p

thanks,
take care.
 
bittersweet said:
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.
 
Hello, bittersweet!


Graph: y  =  x55x4\displaystyle y \;= \;x^5\,-\,5x^4
x-intercepts:   x55x4=0        x4(x5)=0        x=0,  5\displaystyle \;x^5\,-\,5x^4\:=\:0\;\;\Rightarrow\;\;x^4(x\,-\,5)\:=\:0\;\;\Rightarrow\;\;x\:=\:0,\;5

Extrema: 5x420x3=0        5x3(x4)=0        x=0,  4\displaystyle \,5x^4\,-\,20x^3\:=\:0\;\;\Rightarrow\;\;5x^3(x\,-\,4)\:=\:0\;\;\Rightarrow\;\;x\:=\:0,\;4

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

That should be enough to graph the function . . .
Code:
              |
              |                        *
              |
      --------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 (0,0)\displaystyle (0,0), a minimum.

Second derivative: \(\displaystyle \L\,y''\;=\;\frac{2(1\,-\,3x^2)}{(x^2\,+\,1)^3}\)
    \displaystyle \;\;Inflection points: (±33,14)\displaystyle \,\left(\pm\frac{\sqrt{3}}{3},\,\frac{1}{4}\right)

Horizontal asymptote: y=1\displaystyle \,y\,=\,1
Code:
                          |
      - - - - - - - - - - + - - - - - - - - - -
          *               |               *
              *           |           *
                 *        |        *
                   *      |      *
                    *     |     *
                      *   |   *
      --------------------o--------------------
                          |

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