Working with Inverse Graph

[SDPY15]

Hey,

I'm working on an inverse graph for pre-calc 12 and it says i have to graph
h(x)=(g^-1)(X).

My teacher hinted that we have to find the inverse for this, but the only thing I could come up with was the following:

h(x)=(1/g)(x)

Unfortunately I have no clue how to graph this because there is more then one variable and my graphing calc won't accept the "g" value.

Could someone suggest what I have to do to graph this?

 

[srmichael]

Hey,

I'm working on an inverse graph for pre-calc 12 and it says i have to graph
h(x)=(g^-1)(X).

My teacher hinted that we have to find the inverse for this, but the only thing I could come up with was the following:

h(x)=(1/g)(x)

Unfortunately I have no clue how to graph this because there is more then one variable and my graphing calc won't accept the "g" value.

Could someone suggest what I have to do to graph this?

Which one is it? \(\displaystyle h(x)=\frac{1}{g}x \ \ \ OR \ \ \ h(x)=(\frac{1}{g})^x\)

The first equation is a line going through the origin with a slope of 1/g and the other is an exponential function. g, most likely, having not seen how the entire problem is written, is just any constant. It's not clear, however, what you are supposed to do. What exactly does the problem state?

 

[SDPY15]

Um, it would be (1/g) times (x).

The complete questions asks us: Sketch a graph of the following and then fill in the table below:

1: f(x)=3^x

2: g(x)=(3^2x)-3

3: h(x)= g^-1(x)

The table basically asks us to fill in domain, range, asymptotes, y-intercepts and x-intercepts. I got the first to alright, but the third one had me stumped. Right now I'm going through Exponents and Logarithms.

By the way, my teacher also stated that the third equation (h(x)= g^-1(x)) would have an x-intercept but no y-intercept, so that made me think that it has to be either an asymptote or a vertical line.

 

[DrPhil]

Hey,

I'm working on an inverse graph for pre-calc 12 and it says i have to graph
h(x)=(g^-1)(X).

My teacher hinted that we have to find the inverse for this, but the only thing I could come up with was the following:

h(x)=(1/g)(x)

Unfortunately I have no clue how to graph this because there is more then one variable and my graphing calc won't accept the "g" value.

Could someone suggest what I have to do to graph this?

Unfortunately the superscript -1 has a different meaning than reciprocal in this context. It means the inverse function, like square becomes square root. The definition is that the inverse function operating on the direct function returns the argument:
\(\displaystyle g^{-1}[g(x)] = x \)

\(\displaystyle g[g^{-1}(x)] = x \)

Generally for graphing you can interchange the x- and y-axes.

 

[SDPY15]

Unfortunately the superscript -1 has a different meaning than reciprocal in this context. It means the inverse function, like square becomes square root. The definition is that the inverse function operating on the direct function returns the argument:
\(\displaystyle g^{-1}[g(x)] = x \)

\(\displaystyle g[g^{-1}(x)] = x \)

Generally for graphing you can interchange the x- and y-axes.

But I don't have any information to graph so how am I supposed to interchange the values? Also, how did you get g(g^-1(x))=x when the equation asks for h(x)?

 

[DrPhil]

Um, it would be (1/g) times (x).

The complete questions asks us: Sketch a graph of the following and then fill in the table below:

1: f(x)=3^x

2: g(x)=(3^2x)-3

3: h(x)= g^-1(x)

The table basically asks us to fill in domain, range, asymptotes, y-intercepts and x-intercepts. I got the first to alright, but the third one had me stumped. Right now I'm going through Exponents and Logarithms.

By the way, my teacher also stated that the third equation (h(x)= g^-1(x)) would have an x-intercept but no y-intercept, so that made me think that it has to be either an asymptote or a vertical line.

It is helpful to have the complete question.

2. \(\displaystyle y = g(x) = 3^{2x} - 3 \)

....\(\displaystyle x = g^{-1}(y) = \frac{1}{2} \log_3(y + 3) \)

...........\(\displaystyle h(x) = \frac{1}{2} \log_3(x + 3) \)

Find the inverse of g(x), and set h(x) equal to that function.
Take the plot of g(x) and flip it about the line x=y, making the old x-axis vertical and the old y-axis horizontal.

 

[SDPY15]

It is helpful to have the complete question.

2. \(\displaystyle y = g(x) = 3^{2x} - 3 \)

....\(\displaystyle x = g^{-1}(y) = \frac{1}{2} \log_3(y + 3) \)

...........\(\displaystyle h(x) = \frac{1}{2} \log_3(x + 3) \)

Find the inverse of g(x), and set h(x) equal to that function.
Take the plot of g(x) and flip it about the line x=y, making the old x-axis vertical and the old y-axis horizontal.

I don't think i'm following you there. As far as i'm concerned each of the equations were supposed to be three separate graphs.

 

[Bob Brown MSEE]

hint



Pay attention to the horizontal asymptote y=-3
in 3^(2x)-3

 

[DrPhil]

I don't think i'm following you there. As far as i'm concerned each of the equations were supposed to be three separate graphs.View attachment 2809

Yes, three separate graphs, but related by the fact that #3 is the inverse function of #2.