Investigation of irrational functions




First this

To maintain the benefits of HTML:

x >= a , means x is greater than or equal to a

x =< a , means x is smaller than or equal to a

x in V , means x is en element of the set V

Investigation of functions

In the text below we only mention the steps and the intermediate results. The missing calculations are left as an exercise. With attention to relationships and understanding we can avoid excessive calculations.

Function 1

Investigation of de graph of y2 = 18 x2 - x4.

This is not a function. We split the task into two parts.

First we investigate y = sqrt(18x2-x4) and then y = - sqrt(18x2-x4).

So, first y = sqrt(18x2-x4)

The function is even. Thus, it is sufficient to investigate the function for x >= 0.

Especially in the calculation and simplification of y' and y" we'll see the benefits.

  1. Domain

     
      x in domain
    
    <=>   18x2-x4 >= 0
    
    <=>   x2 ( 18 - x2) >= 0
    
                             after examination of the sign
    
    <=>  x in [ -sqrt(18), sqrt(18)]
    
    From y = sqrt(18x2-x4) we see that y >=0.

  2. Intersection points with the axes

     
     (0,0) (sqrt(18),0) (-sqrt(18),0)
    

  3. Asymptotes

    There are no vertical asymptotes. This follows, without any calculation, from investigation of y'. (see below)

    Because of the limited domain, there are no horizontal and no oblique asymptotes.

  4. Investigation of y' with x > 0.
     
           2 (3+x)(3-x)
    y' =  ---------------
          sqrt(18-x2)
    
    The function is rises between 0 and 3 and decreases between 3 and sqrt(18). If x=3, we have a maximum with a horizontal tangent line.

  5. Investigation of y" with x > 0.

     
             2x.(x2-27)
    y" = -----------------------
         ((18-x2). sqrt(18-x2)
    
    Between 0 and sqrt(18) y" is negative. The convex side is upwards.

  6. Investigation in O(0,0)

    If x--> 0 and x > 0 then lim y' = sqrt(18).

    Since the function is even, there is a vertex in O(0,0).

  7. Angle between the two tangent lines in O.

    Between the tangent line with slope sqrt(18) and the x-as, there is a 77 degrees angle. The angle between the two tangent lines is about 26.5 degrees.

  8. y' for x = sqrt(18)
     
    If  x-->  sqrt(18)  and x <  sqrt(18)  then   lim y' = -infty
    
    The tangent line is vertical.

  9. Plot

    Make a plot and verify all the results above.

  10. Transformation

    Now, consider the function y = a.sqrt(18x2-x4) with a > 0. We calculate a value for the parameter a, such that the two tangent lines in O are orthogonal.

    Without further calculations, we see that the tangent lines are y = a sqrt(18) x en y = -a sqrt(18). From this it follows that for a= 1/sqrt(18) the tangent lines are orthogonal.

  11. Smooth curve The graph of y = - sqrt(18x2-x4) is is the mirror image of the graph, relative to the x-axis. If you plot both graphs together, you have a smooth curve. Make the plot!

    y2 = 18 x2 - x4 is the equation of this smooth curve.

  12. Plot

    Make a plot and verify all the results above.

Function 2

y = 0.25 sqrt(x4+ 3 x3)

  1. Domain

    x is in the domain if and only if x4+ 3 x3 >= 0.

    After examination of the sign we see that: domain = (-infty, -3] u [ 0 , +infty)

  2. Intersection points with the axes

    (-3,0) (0,0)

  3. Asymptotes

    There are no vertical asymptotes.

    There are no horizontal asymptotes.

    If x --> +infty then lim f(x)/x = +infty

    If x --> -infty then lim f(x)/x = -infty

    There is no oblique asymptote

  4. Calculation of y'

     
              (4x+ 9) x2
        y' = ----------------------
             8.sqrt(x4+ 3 x3)
    
    After examination of the sign we see that the function is rising in [ 0 , +infty) and decreases in (-infty, -3].

    If x --> 0 lim y' = 0. The curve is tangent to the x-axis.

    If x --> 3 lim y' = -infty. The tangent line in point (-3,0) is vertical.

  5. Calculation of y"

    First we simplify y' to

     
             4x + 9
       y' = --------------
            8.sqrt(1+3/x)
    
    After differentiation and simplification we find if for x > 0
     
             (8x2 + 36x + 27 )
      y" = ---------------------------
            16 (x + 3) sqrt(x2 + 3 x)
    
    The zeros are -0.95 and -3.55. So, for x > 0, there are no inflection points and the curve has its convex side downwards.

    Now, we take x < 0 . Then

     
            - (8x2 + 36x + 27 )
      y" = ---------------------------
            16 (x + 3) sqrt(x2 + 3 x)
    
    The zeros are -0.95 and -3.55. there is one inflection point.

  6. Plot

    Make a plot and verify all the results above.

Function 3

We'll find the x value such that y = 0.25 sqrt( - x4 - 3 x3) attains its maximum.
  1. Domain

    x is in the domain if and only if (- x4 - 3 x3) = - x3(x+3) is not negative.

    With sign investigation we see that the domain is [-3,0].

  2. Maximum

    In the domain [-3,0] we have y > 0. So,

     
         y is maximum  in [-3,0]
    <=>
         16 y2 is maximum in [-3,0]
    <=>
         u = - x4 - 3 x3  is maximum in [-3,0]
    <=>
         u' = -4 x3 -9 x2 = 0  in [-3,0]
    <=>
         -x2 (4x + 9) =  0  in [-3,0]
    <=>
         x = -9/4
    



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