S 4 years ago How to sketch the graph of f(x)=x^2(x^2-4) including the inflection point?

1. anonymous

start with $y=x^2(x-2)(x+2)$ so you see that you have 3 zeros, at x =-2,x=0,x=2

2. anonymous

the zero at x = 0 has multiplicity 2, so the graph will touch the x - axis there but not cross it

3. anonymous

to find the point of inflection you need the second derivative, $f(x)=x^4-4x^2$ $f'(x)=4x^3-8x$ $f''(x)=12x^2-8$ set equal to zero and solve

4. anonymous

My main question is that, when i find the inflection point I put square root of 2/3 into the original formula and i get -20/3 right?

5. anonymous

And this point can not be on the graph, this makes me confused

6. anonymous

yes, for the second coordinate you evaluate the function at $\sqrt{\frac{2}{3}}$

7. anonymous

yes, so I should get 2 inflection points

8. anonymous

it must be on the graph, plug in the x, find the y

9. anonymous

right $\pm\sqrt{\frac{2}{3}}$

10. anonymous

so when I plug I get - 20/3

11. anonymous

for point of inflection we need only to equate first derivative of f(x) to 0.not the second

12. anonymous

actually it is the second first derivative gives relative max or min

13. anonymous

it should be plugged in the original formula, right?

14. anonymous

yes, original function you have the x, you want the y

15. anonymous

correct

16. anonymous

But in the end my inflection point comes up to be (- square root of 2/3 ; - 20/3) and (square root of 2/3 ; - 20/3) and my minimum points are (-2;-4) and (2;-4) and it seems that the point of inflection cant be on the graph

17. anonymous

but im sure that i can get the points of inflection from first derivative,not the second