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moneysotrue

  • 3 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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  1. Denebel
    • 3 years ago
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    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
    • 3 years ago
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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?

  3. Denebel
    • 3 years ago
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    If you set your second derivative equal to 0, you will get a point.

  4. moneysotrue
    • 3 years ago
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    Should I use the quadratic formula?

  5. Denebel
    • 3 years ago
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    Second derivative = 12x + 4 = 0 12x = -4 x= - 1/3

  6. Denebel
    • 3 years ago
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    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
    • 3 years ago
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    No the second derivative would be 12\[x^2\]+4

  8. moneysotrue
    • 3 years ago
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    b/c arent I suppose to subtract 1 from my exponent? each time I take my derivative

  9. Denebel
    • 3 years ago
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    Oops, yes. Oh I see now. I would say no points of inflection

  10. Denebel
    • 3 years ago
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    12x^2 + 4 = 0 x^2 = -1/3 but squares of numbers cannot be negatives

  11. moneysotrue
    • 3 years ago
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    okay thanks! I appreciate the help.

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