## moneysotrue 3 years ago Given: f'(x)=4x^3+4x+6. How many points of inflection will the graph of f(x) have?

1. Denebel

Find the second derivative... find the derivative of this (since this equation is already the first derivative) and set it equal to 0.. that will be where your point(s) of inflection will be.

2. moneysotrue

for my second derivative i got f"(x)=12x^2+4 but I dont think there will be any points am I correct?

3. Denebel

If you set your second derivative equal to 0, you will get a point.

4. moneysotrue

Should I use the quadratic formula?

5. Denebel

Second derivative = 12x + 4 = 0 12x = -4 x= - 1/3

6. Denebel

So you will have 1 point of inflection. You can check this by integrating the first derivative to find the original function. f(x) would be x^4 + 2x + 6x. Graph this and you will see there is one point of inflection.

7. moneysotrue

No the second derivative would be 12\[x^2\]+4

8. moneysotrue

b/c arent I suppose to subtract 1 from my exponent? each time I take my derivative

9. Denebel

Oops, yes. Oh I see now. I would say no points of inflection

10. Denebel

12x^2 + 4 = 0 x^2 = -1/3 but squares of numbers cannot be negatives

11. moneysotrue

okay thanks! I appreciate the help.