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

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

First we investigate y = sqrt(18x^{2}-x^{4}) and then y = - sqrt(18x^{2}-x^{4}).

So, first y = sqrt(18x^{2}-x^{4})

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.

- Domain
x in domain <=> 18x

From y = sqrt(18x^{2}-x^{4}>= 0 <=> x^{2}( 18 - x^{2}) >= 0 after examination of the sign <=> x in [ -sqrt(18), sqrt(18)]^{2}-x^{4}) we see that y >=0. - Intersection points with the axes
(0,0) (sqrt(18),0) (-sqrt(18),0)

- 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.

- Investigation of y' with x > 0.
2 (3+x)(3-x) y' = --------------- sqrt(18-x

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.^{2}) - Investigation of y" with x > 0.
2x.(x

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

- 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.

- y' for x = sqrt(18)
If x--> sqrt(18) and x < sqrt(18) then lim y' = -infty

The tangent line is vertical. - Plot
Make a plot and verify all the results above.

- Transformation
Now, consider the function y = a.sqrt(18x

^{2}-x^{4}) 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.

- Smooth curve
The graph of y = - sqrt(18x
^{2}-x^{4}) 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!y

^{2}= 18 x^{2}- x^{4}is the equation of this smooth curve. - Plot
Make a plot and verify all the results above.

- Domain
x is in the domain if and only if x

^{4}+ 3 x^{3}>= 0.After examination of the sign we see that: domain = (-infty, -3] u [ 0 , +infty)

- Intersection points with the axes
(-3,0) (0,0)

- 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

- Calculation of y'
(4x+ 9) x

After examination of the sign we see that the function is rising in [ 0 , +infty) and decreases in (-infty, -3].^{2}y' = ---------------------- 8.sqrt(x^{4}+ 3 x^{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.

- 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**(8x

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.^{2}+ 36x + 27 ) y" = --------------------------- 16 (x + 3) sqrt(x^{2}+ 3 x)Now, we take

**x < 0**. Then- (8x

The zeros are -0.95 and -3.55. there is one inflection point.^{2}+ 36x + 27 ) y" = --------------------------- 16 (x + 3) sqrt(x^{2}+ 3 x) - Plot
Make a plot and verify all the results above.

- Domain
x is in the domain if and only if (- x

^{4}- 3 x^{3}) = - x^{3}(x+3) is not negative.With sign investigation we see that the domain is [-3,0].

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

y is maximum in [-3,0] <=> 16 y

^{2}is maximum in [-3,0] <=> u = - x^{4}- 3 x^{3}is maximum in [-3,0] <=> u' = -4 x^{3}-9 x^{2}= 0 in [-3,0] <=> -x^{2}(4x + 9) = 0 in [-3,0] <=> x = -9/4

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