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Math Help - factoring to find intervals of a function...

  1. #1
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    factoring to find intervals of a function...

    I have this function : f(x)=x^3-x^2+4x-3

    I need to find the intervals of increase and decrease so i have to factor the first derivative of that function which is f'(x)=3x^2-2x+4

    Now i need to as it says here...put the first derivative equal to zero. But the problem being is that you can't really factor this regularly...so you have to use the quadratic equation. Am I able to use quadratic formulae? for the zeros that I need to find?
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    Re: factoring to find intervals of a function...

    Well, that is what the quadratic formula is for! Yes, you can use the quadratic formula to find the zeros of this quadratic.

    However, this particular polynomial does not have real zeros. I suspect that was why you asked! When x= 0, f'(0)= 4> 0 and f'(x) is never 0. What does that tell you?
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    Re: factoring to find intervals of a function...

    Quote Originally Posted by sara213 View Post
    I have this function : f(x)=x^3-x^2+4x-3

    I need to find the intervals of increase and decrease so i have to factor the first derivative of that function which is f'(x)=3x^2-2x+4

    Now i need to as it says here...put the first derivative equal to zero. But the problem being is that you can't really factor this regularly...so you have to use the quadratic equation. Am I able to use quadratic formulae? for the zeros that I need to find?
    The function is increasing when \displaystyle f'(x) > 0 and decreasing when \displaystyle f'(x) < 0. It is nearly always easiest to solve quadratic inequalities by completing the square.

    \displaystyle \begin{align*}3x^2 - 2x + 4 &= 3\left(x^2 - \frac{2}{3}x + \frac{4}{3}\right) \\ &= 3\left[x^2 - \frac{2}{3}x + \left(-\frac{1}{3}\right)^2 - \left(-\frac{1}{3}\right)^2 + \frac{4}{3}\right] \\ &= 3\left[\left(x - \frac{1}{3}\right)^2 - \frac{1}{9} + \frac{12}{9}\right] \\ &= 3\left[\left(x - \frac{1}{3}\right)^2 + \frac{11}{9}\right] \\ &= 3\left(x - \frac{1}{3}\right)^2 + \frac{11}{3}\end{align*}.

    This is always positive, so the function is increasing for all \displaystyle x.
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    Re: factoring to find intervals of a function...

    @hallofIvy

    so it is always increasing?
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    Re: factoring to find intervals of a function...

    Quote Originally Posted by sara213 View Post
    @hallofIvy

    so it is always increasing?
    Yes, and there are not any stationary points since \displaystyle f'(x) \neq 0 for any \displaystyle x.
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    Re: factoring to find intervals of a function...

    This is still the same question that i'm working on.


    Is it fair to say that this function is always increasing? and that is has no concave up or concave down feature because that shape must look in the shape of u correct? how would i know exactly if the graph didn't have a concave down or up? also can it still have an inflection point if it is always increasing?

    I'm sorry if they seem silly questions.....:S
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    Re: factoring to find intervals of a function...

    Quote Originally Posted by sara213 View Post
    This is still the same question that i'm working on.


    Is it fair to say that this function is always increasing? and that is has no concave up or concave down feature because that shape must look in the shape of u correct? how would i know exactly if the graph didn't have a concave down or up? also can it still have an inflection point if it is always increasing?

    I'm sorry if they seem silly questions.....:S
    As it has been shown that no matter the value of x, the derivative is always positive, yes it is fair to say the function is always increasing.

    The concavity of the function is determined by the second derivative. It is concave down where the second derivative is negative, and concave up where the second derivative is positive.
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