Questions on Algebra: Functions, Domain, NOT graphing answered by real tutors!

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Question 166656: I missed the day of class when we covered domains in my class, and now i am hopelessly lost.
Please help me solve the following problems, and explain how you did it. I am not very good at math, so the more you can break it down and explain it, the better.
find the domain of the rational function:
h(t)=(t-5)/(t^2-25)

g(x)=(x^3-27)/(4x)

the problems are written in faction form, but I don't know how to do that with the computer, so I am not sure if what i have written means the same thing. It should read t-5 over t^2-25.
I would really appreciate the help!: I missed the day of class when we covered domains in my class, and now i am hopelessly lost.
Please help me solve the following problems, and explain how you did it. I am not very good at math, so the more you can break it down and explain it, the better.
find the domain of the rational function:
h(t)=(t-5)/(t^2-25)

g(x)=(x^3-27)/(4x)

the problems are written in faction form, but I don't know how to do that with the computer, so I am not sure if what i have written means the same thing. It should read t-5 over t^2-25.
I would really appreciate the help!
Answer by jim_thompson5910(9869) About Me  (Show Source):
You can put this solution on YOUR website!
Remember, the domain is simply the set of all x values that produce a y value.

# 1



h(t)=(t-5)/(t^2-25) Start with the given function


t^2-25=0 Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of "t" that make the denominator zero, then we must exclude them from the domain.




(t-5)(t+5)=0 Factor the left side (note: if you need help with factoring, check out this solver)




Now set each factor equal to zero:

t-5=0 or t+5=0

t=5 or t=-5 Now solve for t in each case


So our solutions are t=5 or t=-5



Since t=-5 and t=5 make the denominator equal to zero, this means we must exclude t=-5 and t=5 from our domain

So our domain is:

which in plain English reads: t is the set of all real numbers except t CANNOT equal -5 or t CANNOT equal 5

So our domain looks like this in interval notation


note: remember, the parenthesis excludes -5 and 5 from the domain



If we wanted to graph the domain on a number line, we would get:

drawing(500,50,-10,10,-10,10,<BR> number_line( 500, -10, 10),<BR> blue(line(-4.5,-7,4.65,-7)),<BR> blue(line(-4.5,-6,4.65,-6)),<BR> blue(line(-4.5,-5,4.65,-5)),<BR> blue(arrow(5.5,-7,10,-7)),<BR> blue(arrow(5.5,-6.5,10,-6.5)),<BR> blue(arrow(5.5,-6,10,-6)),<BR> blue(arrow(5.5,-5.5,10,-5.5)),<BR> blue(arrow(5.5,-5,10,-5)),<BR> blue(arrow(-5.5,-7,-10,-7)),<BR> blue(arrow(-5.5,-6.5,-10,-6.5)),<BR> blue(arrow(-5.5,-6,-10,-6)),<BR> blue(arrow(-5.5,-5.5,-10,-5.5)),<BR> blue(arrow(-5.5,-5,-10,-5)), <BR> circle(-5,-5.8,0.35),<BR> circle(-5,-5.8,0.4),<BR> circle(-5,-5.8,0.45), <BR> <BR> circle(5,-5.8,0.35),<BR> circle(5,-5.8,0.4),<BR> circle(5,-5.8,0.45) <BR> <BR> ) Graph of the domain in blue and the excluded values represented by open circles

Notice we have a continuous line until we get to the holes at t=-5 and t=5 (which is represented by the open circles).
This graphically represents our domain in which t can be any number except t cannot equal -5 or 5







# 2





g(x)=(x^3-27)/(4x) Start with the given function


4x=0 Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of x that make the denominator zero, then we must exclude them from the domain.



x=(0)/(4) Divide both sides by 4 to isolate x



x=0 Divide





Since x=0 makes the denominator equal to zero, this means we must exclude x=0 from our domain

So our domain is:

which in plain English reads: x is the set of all real numbers except x CANNOT equal 0

So our domain looks like this in interval notation


note: remember, the parenthesis excludes 0 from the domain

If we wanted to graph the domain on a number line, we would get:

drawing(500,50,-10,10,-10,10,<BR> number_line( 500, -10, 10),<BR> blue(arrow(0.2,-7,10,-7)),<BR> blue(arrow(0.2,-6.5,10,-6.5)),<BR> blue(arrow(0.2,-6,10,-6)),<BR> blue(arrow(0.2,-5.5,10,-5.5)),<BR> blue(arrow(0.2,-5,10,-5)),<BR> blue(arrow(-0.2,-7,-10,-7)),<BR> blue(arrow(-0.2,-6.5,-10,-6.5)),<BR> blue(arrow(-0.2,-6,-10,-6)),<BR> blue(arrow(-0.2,-5.5,-10,-5.5)),<BR> blue(arrow(-0.2,-5,-10,-5)), <BR> circle(0,-5.8,0.35),<BR> circle(0,-5.8,0.4),<BR> circle(0,-5.8,0.45),<BR> circle(0,-5.8,0.4),<BR> circle(0,-5.8,0.45)<BR> ) Graph of the domain in blue and the excluded value represented by open circle

Notice we have a continuous line until we get to the hole at x=0 (which is represented by the open circle).
This graphically represents our domain in which x can be any number except x cannot equal 0