SOLUTION: Let {{{f(x)=7/(5-x)}}} (i) State the domain. (ii) State the vertical asymptote(s). (iii) State the horizontal asymptote. (iv) Find the inverse fu

Algebra ->  Graphs -> SOLUTION: Let {{{f(x)=7/(5-x)}}} (i) State the domain. (ii) State the vertical asymptote(s). (iii) State the horizontal asymptote. (iv) Find the inverse fu      Log On


   

Question 150025: Let f%28x%29=7%2F%285-x%29


(i) State the domain.


(ii) State the vertical asymptote(s).


(iii) State the horizontal asymptote.



(iv) Find the inverse function of f. Show work.




Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

i)



f%28x%29=%287%29%2F%285-x%29 Start with the given function


5-x=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=0-5Subtract 5 from both sides


-x=-5 Combine like terms on the right side


x=%28-5%29%2F%28-1%29 Divide both sides by -1 to isolate x



x=5 Divide





Since x=5 makes the denominator equal to zero, this means we must exclude x=5 from our domain

So our domain is:

which in plain English reads: x is the set of all real numbers except x%3C%3E5

So our domain looks like this in interval notation


note: remember, the parenthesis excludes 5 from the domain

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ii)
Vertical Asymptote:
To find the vertical asymptote, just set the denominator equal to zero and solve for x

5-x=0 Set the denominator equal to zero


-x=0-5Subtract 5 from both sides


-x=-5 Combine like terms on the right side


x=%28-5%29%2F%28-1%29 Divide both sides by -1 to isolate x



x=5 Divide


So the vertical asymptote is x=5

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iii)

Looking at the numerator 7, we can see that the degree is 0 since the highest exponent of the numerator is 0. For the denominator 5-x, we can see that the degree is 1 since the highest exponent of the denominator is 1.


Horizontal Asymptote:

Since the degree of the numerator (which is 0) is less than the degree of the denominator (which is 1), the horizontal asymptote is always y=0

So the horizontal asymptote is y=0



Notice if we graph y=%287%29%2F%285-x%29, we can visually verify our answers:

Graph of y=%287%29%2F%285-x%29%29 with the horizontal asymptote y=0 (blue line) and the vertical asymptote x=5 (green line)

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iv)

f%28x%29=7%2F%285-x%29 Start with the given function.


x=7%2F%285-f%28x%29%29 Switch x and f(x)


x%285-f%28x%29%29=7 Multiply both sides by 5-f%28x%29.


5x-x%2Af%28x%29=7 Distribute


-x%2Af%28x%29=7-5x Subtract 5x from both sides.


f%28x%29=%287-5x%29%2F%28-x%29 Divide both sides by -x.


f%28x%29=%28-7%2B5x%29%2F%28x%29 Reduce.


f%28x%29=%285x-7%29%2F%28x%29 Rearrange the terms.


So the inverse function is