Inverse of a Function

[jpnov]

Really not sure where to go with this. Not sure where to start or what I need to do. I understand that shifting the curve to the left will shift it farther below the line y=x when reflected about it. Is that a good explanation for that part of (a)??

How would I find an expression for the inverse of g(x) = f(x+c)?

And how would i go about solving part b?

Thanks

 

[mmm4444bot]

I understand that shifting the curve to the left will shift it farther below the line y=x when reflected about it. Is that a good explanation for that part of (a)?

You've got some pronoun issues, but what you're thinking is a good start. You just need a better explanation.

Rather than saying, "farther below the line y = x", I would simply say that shifting a function's graph to the left causes the graph of its inverse to shift down.

After all, if the graph of the inverse is above the line y=x to begin with, shifting the other function to the left might not lower the inverse graph enough to even reach the line y=x, let alone fall below it.


How would I find an expression for the inverse of g(x) = f(x+c) ?

How about some "analysis" ?

I would hope that you've first determined inverses of various functions, before receiving this exercise.

So, you could experiment with a couple of concrete-example functions, and look for a "pattern".


And how would i go about solving part b ?

I would go about it in the same way as part (a); search for a deeper understanding of what's happening ("analysis"). Experiment with some example functions. Determine their inverses. Compare graphs of both, before and after shifting.

Then, looking at your exercises, try to think "symbolically".

Here's an easy example of some function f(x) and its inverse f[sup:1ee87we5]-1[/sup:1ee87we5](x).

f(x) = ln(x - 4)

Now, if we shift the graph of f to the left four units, then we get a new function:

g(x) = ln(x).

In other "symbolic" words, g(x) = f(x + 4).

If we shift another four units to the left, we get another new function:

h(x) = ln(x + 4)

In other "symbolic" words, h(x) = g(x + 4).

The inverses in this progression are:

f[sup:1ee87we5]-1[/sup:1ee87we5](x) = e^x + 4

g[sup:1ee87we5]-1[/sup:1ee87we5](x) = e^x

h[sup:1ee87we5]-1[/sup:1ee87we5](x) = e^x - 4

So we "see" that the inverse function shifts downward four units each time ITS inverse shifts to the left four units.

Does this concrete example prompt any thoughts about part (a) ?

Does the strategy give you any ideas about what to try for part (b) ?

Here are graphs of the progression in my example. (Well, almost the graphs. These three are close enough to show the relative shifts, which is my only intent.)

f(x), g(x), and h(x) are plotted in green.

Their inverses are plotted in red.

The line y = x is plotted in yellow.

As f shifts to the left c units, its inverse shifts down c units, where c = 4.

[attachment=2:1ee87we5]ln(x - 4).JPG[/attachment:1ee87we5]

[attachment=1:1ee87we5]ln(x).JPG[/attachment:1ee87we5]

[attachment=0:1ee87we5]ln(x+4).JPG[/attachment:1ee87we5]

 

[jpnov]

Thank you for your help. For part a, I figure that the inverse of g(x) = f(x+c) is g^(-1)(x) = f^(-1)(x) + c. Is this correct? I'm working on part b now. Thanks again!

And for part b, is the answer h^(-1)(x) = [f^(-1)]/c ?
I'm fairly confused on this part

 

[mmm4444bot]

the inverse of g(x) = f(x+c) is

g^(-1)(x) = f^(-1)(x) + c.

Is this correct? Almost.

Be careful when working with a symbolic contant, when the sign of that unknown number is crucial to building a proper expression.

It comes back to the age-old "tricky" question: "Is -x a positive number or a negative number?"

The correct answer is, "We can't say; it depends upon the sign of the value represented by the symbol x."

In other words, they gave you f(x + c), and they told you that f shifted to the left c units. Therefore, in this exercise, c must represent a positive value, yes?

I mean, if c were to be a negative number, then x + c represents subtraction from x, and that means a shift to the right, not left.

Now go back and examine your answer for g[sup:1863rizd]-1[/sup:1863rizd](x) -- knowing that the contant c is a positive number, in this exercise.


for part b, is the answer h^(-1)(x) = [f^(-1)]/c

Exactly !

The inverse operation of multiplication is division.

Or, said another way, if we multiply x by a non-zero constant c, then we "undo" that multiplication with division by c.

f(x) = 1/2 x

Let c = 2

g(x) = f(cx) = x

To get from x back to 1/2 x we "inverse" the change.

From a graphical perspective, multiplying x by positive c vertically stretches the graph; to get back, divison by c vertically compresses the graph of f inverse.

g[sup:1863rizd]-1[/sup:1863rizd](x) = 1/c * f[sup:1863rizd]-1[/sup:1863rizd](x) = 1/2(x) = 1/2 x