How do you find intercepts, extrema, points of inflections, asymptotes and graph #y=(20x)/(x^2+1)-1/x#?

1 Answer
Nov 14, 2016

There is a vertical asymptote at #x=0#
There is a horizontal asymptote as #x rarr +- oo # along the #x#-axis
x-intercepts when #x=+-1/sqrt19#
y(x) is an odd function, and it is symmetric about the origin O

Critical points are:
Maximum at #(1.096, 9.046)#
Minimum at #(-1.096, -9.046)#

Explanation:

# y=(20x)/(x^2+1)-1/x #
# :. y=((20x^2)-(x^2+1))/(x(x^2+1)) #
# :. y=(19x^2-1)/(x(x^2+1)) #

Vertical Asymptotes
These will occur when the denominator is zero

# :. x(x^2+1)) =0 => x=0 #

So There is a vertical asymptote at #x=0#

Horizontal Asymptotes
We need to examine the behaviour as #x rarr +- oo#.
For large #x#,
# 19x^2-1 ~ x^2 #
# x^2+1 ~ x^2 #
So, # (19x^2-1)/(x(x^2+1)) ~ 19x^2/x^3 ~ 1/x ->0# as #x rarr +- oo #
So There is a horizontal asymptote as #x rarr +- oo # along the #x#-axis

Odd/Even Function
Find y(-x);

# y(-x)=(19(-x)^2-1)/(-x((-x)^2+1)) #
# y(-x)=(19x^2-1)/(-x(x^2+1)) #
# y(-x)=-(19x^2-1)/(x(x^2+1)) #
# y(-x)=-y(x) #

Hence y(x) is an odd function, and it is symmetric about the origin O

Intercepts

# y=(19x^2-1)/(x(x^2+1)) => 19x^2-1 = 0 #
# :. x^2=1/19 #
# :. x^2=+-1/sqrt19 ~~~ +- 0.229#

We have already established that x=0 is a vertical asymptotes so there are not y-axis intercepts.
Hence x-intercepts when #x=+-1/sqrt19#

Summary
There is a vertical asymptote at #x=0#
There is a horizontal asymptote as #x rarr +- oo #
y(x) is an odd function, and it is symmetric about the origin O

Extrema
Because of symmetry we only need to identify extra for #x>=0#
We can write;

# y=(19x^2-1)/(x^3+x) #

Using the quotient rule

# y=(19x^2-1)/(x^3+x) #
# dy/dx=((x^3+x)(d/dx(19x^2-1)) - (19x^2-1)(d/dx(x^3+x)) )/(x^3+x)^2 #
# :. dy/dx=((x^3+x)(38x) - (19x^2-1)(3x^2+1) )/(x^3+x)^2 #
# :. dy/dx=(38x^4+38x^2 - (57x^4+16x^2-1) )/(x^3+x)^2 #
# :. dy/dx=(38x^4+38x^2 - 57x^4-16x^2+1 )/(x^3+x)^2 #
# :. dy/dx=(-19x^4+22x^2 +1 )/(x^3+x)^2 #
# :. dy/dx=-((19x^4-22x^2 -1 ))/(x^3+x)^2 #

At extrema, # dy/dx=0 => 19x^4-22x^2 -1 =0#
This is quadratic in #x^2# and can be solved, which I will return to!
To solve this we use the quadratic formula

# x=(-b+-sqrt(b^2-4ac))/(2a)#

to find two solutions for #x^2#, and then take their square roots, which yields two real solutions;

# x=+-1.096 # (3dp)

Due to the symmetry about O, we only need to examine one of these critical points as the other will be the opposite. Let's look at #x=1.096#

By inspection (using a calculator with #x=1.096+-epsilon#, where #epsilon# is small we can see this point corresponds to a maximum, and so #x=-1.096# is a minimum. We could also investigate using the second derivative test, but due to the complexity a simple computed test is easier.

When #x=1.096 => y=9.046# (3dp)

Hence, Critical points are:
Maximum at #(1.096, 9.046)#
Minimum at #(-1.096, -9.046)#

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