How do you find the domain and range of #f(x)=(12x)/(x^2-36)#?

Question

Ethan Adkins · Accepted Answer

Below

Looking at the graph, you can immediately see that there are 2 vertical asymptotes because #x^2-36 !=0# so #x=+-6# are the vertical asymptotes. Therefore, the graph cannot have the points with the x-coordinates #x=6# and #x=-6# The horizontal asymptote is #y=0# since the degree of the numerator is less than the degree of the denominator. This is because if you imagine letting #x# be any number, then #x^2-36# will be a whole lot bigger than #12x# and since the small number divided by a larger number, then #(12x)/(x^2-36) ->0# Therefore, the graph cannot have the points with the y-coordinate #y=0# But what asymptotes actually tell you about the graph is that its end points will approach the horizontal and vertical asymptotes, but they will never actually touch them. In other words, asymptotes describe the graph's shape, which can help you identify the graph's domain and range. Intercepts When #y=0#, #x=0# When #x=0#, #y=0# You will notice that the graph can pass through #(0,0)# but the endpoints of the graph will be approaching #y=0# and not cross #y=0#. This is because asymptotes affect the end points only. Hence, Domain: all reals #x# except when #x=+-6# Range: all reals #y# The graph is shown below. graph{(12x)/(x^2-36) [-10, 10, 5, 5, 10]}

Genesis Allen · Answer

The domain is #x in (-oo, -6)uu-6,6)uu(6,+oo)#. The range is #y in RR#

The denominator must be #!=0#

Consequently,

#x^2-36!=0#

#(x+6)(x-6)!=0#

#x!=-6# and #x!=6#

The domain is #x in (-oo, -6)uu-6,6)uu(6,+oo)#

To determine the range, let

#y=(12x)/(x^2-36)#

#y(x^2-36)=12x#

#yx^2-12x-36y=0#

This is a quadratic equation in #x# and in order to have solutions, the discriminant #>=0#

Consequently,

#Delta=(-12)^2-4(y)(-36y)#

#=144+144y^2>=0#

#=>#, #144(1+y^2)>=0#

Consequently,

#AA y in RR, Delta>0#

The range is #y in RR#

graph{12x/(x^2-36)[-16.24, 16.25, 32.49, 32.46]}

Isaiah Anthony · Answer

undefined To find the domain of the function $ f(x) = \frac{12x}{x^2 - 36} $, we need to identify any values of $ x $ that would make the denominator equal to zero. In this case, the denominator is $ x^2 - 36 $, so we set it equal to zero and solve for $ x $. The solutions are $ x = -6 $ and $ x = 6 $. Therefore, the domain of the function is all real numbers except $ x = -6 $ and $ x = 6 $.

To find the range of the function, we consider the behavior of the function as $ x $ approaches positive or negative infinity. As $ x $ approaches positive or negative infinity, the function approaches zero. Therefore, the range of the function is all real numbers except zero.