Anonymous asked in Science & MathematicsMathematics · 1 month ago

How to find the stationary points for (x-3)(2x-1)?

I have found derivative 4x-5,substituted and got 5/4

4 Answers

  • 1 month ago
    Favorite Answer

    There are two ways we can get the derivative.  We can expand the product of two binomials into a quadratic and then get the derivative or we can use the product rule.  Either way, we should get the same result.  I'll use the product rule:

    y = (x - 3)(2x - 1)

    y = uv and u = x - 3 and v = 2x - 1

    y' = u'v + uv' and u' = 1 and v' = 2

    This gives us:

    y' = 1(2x - 1) + (x - 3)(2)

    y' = 2x - 1 + 2x - 6

    y' = 4x - 7

    This was where your problem was.  You were close but didn't quite get the correct answer.

    From here, we solve this expression for its zero to get the "x" value of the original function for the stationary point, you did this (with the incorrect expression) to get an answer, so the step was right, just with the wrong info:

    0 = 4x - 7

    7 = 4x

    7/4 = x

    That's just the "x" value of the point.  We now put that into the original equation to get the y:

    y = (x - 3)(2x - 1)

    y = (7/4 - 3)(2 * 7/4 - 1)

    y = (7/4 - 12/4)(7/2 - 2/2)

    y = (-5/4)(5/2)

    y = -25/8

    The point of the stationary point is:

    (7/4, -25/8)

    Attachment image
  • Vaman
    Lv 7
    1 month ago

    For stable equilibrium point, the first derivative should be 0 and the second derivative should be positive. y= (x-3)(2x-1), dy/dx= (2x-1)+2(x-3)=0


     x=7/4, The second derivative is 4. It is positive. x =7/4 is the equilibrium point. Y value at x=7/4 is


    The point is (7/4,-25/8)

  • 1 month ago

    Double-check your math. I don't think you expanded correctly before taking the first derivative. The 4x is correct, but the -5 is not.

    Set the revised expression equal to zero and solve for x.

    In order to find y, you just plug in that x value into the original function.

  • rotchm
    Lv 7
    1 month ago

    Hint: First, you need to know what stationary points mean. So look in your notes and state it here. Then unanon yourself and we will take it from there.

    Hopefully no one will spoil you the answer. That would be very irresponsible of them.

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