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Thread: Differentiation

  1. #1
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    Question Differentiation

    f(x)=x^3-3x^2+3x
    f '(x)=3x^2-6x+3
    I have no issue with finding f ' (x).

    The next part of the question says, thus show that f' (x) => 0 for all x and interpret this in terms of the graph f(x)=x^3-3x^2+3x

    If anyone can help that would be great.
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  2. #2
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    Quote Originally Posted by thekiterunner View Post
    f(x)=x^3-3x^2+3x
    f '(x)=3x^2-6x+3
    I have no issue with finding f ' (x).

    The next part of the question says, thus show that f' (x) => 0 for all x and interpret this in terms of the graph f(x)=x^3-3x^2+3x

    If anyone can help that would be great.
    Simply solve the inequality $\displaystyle 3x^2-6x+3 \geq 0 $ by using the fact that $\displaystyle 3x^2-6x+3 = 3(x^2-2x+1)$.

    After you proved that, you know $\displaystyle f'(x) \geq 0 \ \forall x \in \mathbb{R}$. What does this tell you about the $\displaystyle f(x)$? What, in general, does $\displaystyle f'(x)$ mean in regards to $\displaystyle f(x)$?
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  3. #3
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    i do not undertsand that one little bit...any chance for simplier terms
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  4. #4
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    If you have a general quadratic equation:
    $\displaystyle f(x)=ax^2+bx+c$

    And let the discriminant $\displaystyle D=b^2-4\cdot ac$.

    Now if D<0 that means the equation has no real solution, that is the graph nowhere goes
    through the x-axis. It is entirely contained on either side. Either $\displaystyle f<0$ or $\displaystyle f>0$ for all x.
    Now if you dont know if f is below or above the x-axis,
    just check for some point if it is above or below, then you know that the rest of the points
    are on the same side.

    If D=0 then there is one real solution, the graph is entirely on one side and "touches" the
    x-axis in only one point, it never crosses the axis. So either $\displaystyle f\leq0$ or $\displaystyle f\geq0$ for all x.


    What if D>0 ?

    But anyway what is D in your equation?


    After you have found out the first part, just think about, like Defunkt said, what $\displaystyle f'(x)$ means in regard to $\displaystyle f(x)$.


    Does that help?
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  5. #5
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    But anyway what is D in your equation?

    D=6^2 - (4 x 3 x 3) = 0

    So becuase D=0 i have one real solution

    f'(x) is just the gradient function of f(x)
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  6. #6
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    So if f'(x) is always positive or zero, then f(x) must be increasing (not strictly).
    And f(x) does not have a local maxima or minima at it's critical point.
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  7. #7
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    The geometrical meaning of f'(x) regarding f(x) is the slope -- f'(x0) is (if it exists) the slope of the function f in point x0.

    Now, if $\displaystyle \forall x \in \mathbb{R}, \ f'(x) \geq 0 $, what does this tell you about the graph of f?
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