Here's the question you clicked on:
moneysotrue
Given: f'(x)=4x^3+4x+6. How many points of inflection will the graph of f(x) have?
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.
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
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.
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.