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Thread: [SOLVED] Differentiation Question

  1. #1
    Member looi76's Avatar
    Jan 2008

    [SOLVED] Differentiation Question

    The equation of a curve is $\displaystyle y = 6x^2 - x^3$. Find the coordinates of the two stationary points on the curve, and determine the nature of each of these stationary points.

    State the set of values of $\displaystyle x$ which $\displaystyle 6x^2 - x^3$ is a decreasing function of $\displaystyle x$.

    The gradient at the point $\displaystyle M$ on the curve is$\displaystyle 12$. Find the equation of the tangent to the curve at $\displaystyle M$.


    $\displaystyle y = 6x^2 - x^3$

    $\displaystyle \frac{dy}{dx} = 6\times2^{2-1} - 3x^{3-1}$

    $\displaystyle \frac{dy}{dx} = 12x - 3x^2$

    $\displaystyle \frac{dy}{dx} = 0$

    $\displaystyle 12x - 3x^2$
    $\displaystyle 3x(4 - x)$

    $\displaystyle 3x = 0$
    $\displaystyle x = 0$

    $\displaystyle 4 - x = 0$
    $\displaystyle -x = -4$
    $\displaystyle x = 4$

    $\displaystyle x = 0$ and $\displaystyle x = 4$

    $\displaystyle y = 6(0)^2 - (0)^3 = 0$
    $\displaystyle y = 6(4)^2 - (4)^3 = 32$

    So, the two coordinates are $\displaystyle (0,0)$ and $\displaystyle (4,32)$.

    How to find the decreasing function of $\displaystyle x$ ?
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  2. #2
    Moo is offline
    A Cute Angle Moo's Avatar
    Mar 2008
    P(I'm here)=1/3, P(I'm there)=t+1/3

    Perfect thing for the first question !

    For the decreasing : a function decreases when its derivative is negative.

    Thus, you have to solve for $\displaystyle 3x(4-x)<0$.

    A quick way to do is to say that it's negative if and only if the signs for 3x and 4-x are different, that is to say that if 3x is positive, then 4-x has to be negative, and if 3x is negative, 4-x has to be positive.

    Hence the solutions are :

    $\displaystyle \left(3x>0 \text{ AND } 4-x<0\right) {\color{red}\text{ or }} \left(3x<0 \text{ AND } 4-x>0\right)$

    This should give you the set of values for x that you want
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  3. #3
    Member Danshader's Avatar
    Mar 2008
    From now stop asking me where is malaysia...
    well another method is to actually find the gradient right before and after the stationary points. Negative value will mean the curve is moving downwards while positive means moving upwards.

    just substitute an x value right after and before the stationary point to obtain y and compare it to the coordinate of the stationary point too see whether the value of y is increasing or decreasing
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