Question #6c313

1 Answer
Jan 16, 2018

Please see below.

Explanation:

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A function is monotonic if it is either entirely non-increasing or non-decreasing. To find out, we need to take the derivative of it:

#f(x)=x^3/(3+x)^2#

We use the Quotient Rule to take its derivative:

#f'(x)=((3+x)^2(3x^2)-x^3(2(3+x)))/(3+x)^4#

#f'(x)=(x^2(x+9))/(3+x)^3#

Let's set this derivative equal to zero and solve for its roots:

#x^2(x+9)=0#

#x=0, -9#

These are the critical points of the function. If we evaluate the derivative function at values to the left and right of each root we find that at #x < -9# the derivative is positive and at #x > -9# the derivative is negative.

This indicates that at #x=9# the function has a maximum.

At #x < 0#, the derivative is positive and at #x > 0#, the derivative is also positive. This indicates that the function is increasing before #x=0# and continues to increase after #x=0#.

Setting the function equal to zero and solving for its roots gives us

#x^3=0# which means we have #x=0# three times. This is called a root with multiplicity of #3#. When a function has a root with an odd multiplicity it crosses the #x#-axis with a flex (slight shift). It does not produce a maximum or a minimum at that point.

Furthermore, at #x=0# the function is #0# but at #x=-9# the function is #=-81/4#.

This indicates that at #x=-9# the maximum we identified is not an absolute maximum but rather a local maximum. The function does not have any absolute maximum or minimum because as #xrarroo# the function #f(x)=y rarroo#. And as #xrarr-oo# the function also goes to #-oo#.

As a result of the fact that the derivative of the function changes signs, the function is not monotone.

To find the asymptotes of the function we have to consider that it is a rational function. Therefore, we set the denominator equal to zero and solve for its roots:

#(3+x)^2=0#

#x=-3# is a vertical asymptote.

We also notice that the numerator is of degree #3# which is a higher degree than degree #2# of the denominator. This means that function does not have a horizontal asymptote but it has a slant one.

To find it, we do a long division to divide the numerator by the denominator which gives us:

#x^3/(3+x)^2=x^3/(x^2+6x+9)=x-6+(27(x+2))/(3+x)^2#

The slant asymptote has the equation:

#y=x-6#.

Here is the graph of the function where you can see the critical points, asymptotes, and the behavior of the function as described above:

enter image source here