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

1. satellite73 Group Title

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 Group Title

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

3. satellite73 Group Title

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 Group Title

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 Group Title

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

6. satellite73 Group Title

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

7. S Group Title

yes, so I should get 2 inflection points

8. satellite73 Group Title

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

9. satellite73 Group Title

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

10. S Group Title

so when I plug I get - 20/3

11. fazna Group Title

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

12. satellite73 Group Title

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

13. S Group Title

it should be plugged in the original formula, right?

14. satellite73 Group Title

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

15. S Group Title

correct

16. S Group Title

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 Group Title

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