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What is the inverse of h?
h(x) = 6x + 1

 

 

A)h-1(x) = 6x - 1

B)h-1(x) = 6x + 1

C)h-1(x) = x/6 -1

D)h-1(x) = 1/6(x - 1)

 

 

What are the vertical asymptotes of R(x) = 3x - 3/x2 - 4?

 

 

A)x = 0

B)x = 1, and x = -1

C)x = 2, and x = -2

D)x = 3, and x = -3

Guest Aug 15, 2017
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2+0 Answers

 #1
avatar+76160 
+1

 

h(x)  = 6x + 1     we can write

 

y  =  6x + 1         the object  is to get x by itself and then "swap" x and y

 

Subtract 1 from both sides

 

y - 1  = 6x    divide both sides by 6

 

[ y - 1 ] / 6   =  x       "swap x and y

 

[x - 1 ] / 6  = y         for y, write  h-1(x)

 

[ x - 1 ] / 6  =    h-1 (x)         and this is the inverse

 

 

What are the vertical asymptotes of R(x) = 3x - 3/(x2) - 4?

 

As shown here, the vertcal asymptote occurs at x  = 0

 

https://www.desmos.com/calculator/uyregpotef

 

 

 

cool cool cool

CPhill  Aug 15, 2017
edited by CPhill  Aug 15, 2017
 #2
avatar+1119 
+1

In the second question, Cphill interpreted it correctly, but if you meant \(R(x)=\frac{3x-3}{x^2-4}\) or \(R(x)=3x-\frac{3}{x^2-4}\), the answer is different. Let's think about what we have to do to figure out the vertical asymptote.

 

In both cases, we have rational fractions. Of course, we can't have a denominator of 0. This means that if we plug in a value for x that results in a denominator that equals 0, then it is officially outside of the domain. Let's figure out when x^2-4=0.

 

\(x^2-4=0\) You might notice that x^2-4 is a difference of 2 squares, but we don't need to take advantage of this, actually. This is because we have no b-term. Add4 to both sides.
\(x^2=4\) Take the square root of both sides.
\(x=\pm2\)  
   

 

This means that the vertical asymptote is at \(x=\pm2\), which corresponds to the answer choice of C.

TheXSquaredFactor  Aug 15, 2017

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