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

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

Should I use the quadratic formula?

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

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

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

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

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

okay thanks! I appreciate the help.