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

#1**+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/(x^{2}) - 4?

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

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

CPhill Aug 15, 2017

#2**+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