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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Find 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for my second derivative i got f"(x)=12x^2+4 but I dont think there will be any points am I correct?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you set your second derivative equal to 0, you will get a point.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Should I use the quadratic formula?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Second derivative = 12x + 4 = 0 12x = 4 x=  1/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No the second derivative would be 12\[x^2\]+4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0b/c arent I suppose to subtract 1 from my exponent? each time I take my derivative

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oops, yes. Oh I see now. I would say no points of inflection

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.012x^2 + 4 = 0 x^2 = 1/3 but squares of numbers cannot be negatives

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay thanks! I appreciate the help.
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