## S 3 years ago How to sketch the graph of f(x)=x^2(x^2-4) including the inflection point?

1. satellite73

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. satellite73

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

3. satellite73

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. S

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. S

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

6. satellite73

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

7. S

yes, so I should get 2 inflection points

8. satellite73

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

9. satellite73

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

10. S

so when I plug I get - 20/3

11. fazna

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

12. satellite73

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

13. S

it should be plugged in the original formula, right?

14. satellite73

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

15. S

correct

16. S

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. fazna

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