anonymous
  • anonymous
Given: f'(x)=4x^3+4x+6. How many points of inflection will the graph of f(x) have?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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.
anonymous
  • anonymous
for my second derivative i got f"(x)=12x^2+4 but I dont think there will be any points am I correct?
anonymous
  • anonymous
If you set your second derivative equal to 0, you will get a point.

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anonymous
  • anonymous
Should I use the quadratic formula?
anonymous
  • anonymous
Second derivative = 12x + 4 = 0 12x = -4 x= - 1/3
anonymous
  • anonymous
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.
anonymous
  • anonymous
No the second derivative would be 12\[x^2\]+4
anonymous
  • anonymous
b/c arent I suppose to subtract 1 from my exponent? each time I take my derivative
anonymous
  • anonymous
Oops, yes. Oh I see now. I would say no points of inflection
anonymous
  • anonymous
12x^2 + 4 = 0 x^2 = -1/3 but squares of numbers cannot be negatives
anonymous
  • anonymous
okay thanks! I appreciate the help.

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