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.
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.
First derivative: \(\displaystyle \L\,y'\;=\;\frac{(x^2\,+\,1)\cdot2x\,-\,x^2\cdot2x}{(x^2\,+\,1)^2}\;=\;\frac{2x}{(x^2\,+\,1)^2}\) The only critical point is (0,0), a minimum.
Second derivative: \(\displaystyle \L\,y''\;=\;\frac{2(1\,-\,3x^2)}{(x^2\,+\,1)^3}\) Inflection points: (±33,41)
"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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