# If f(x) = x + cos x, find the inverse of f(1)?

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#### ram1cal

##### New Member
if it makes things clearer, it didn't say inverse, it showed f^-1(1), which pretty much means f prime

please show steps? this is the only weird one on my homework

#### MrAl

##### Well-Known Member
Hi,

f(x)=x+cos(x)

and it looks like you need to find f^-1(1) which would mean if we
had the inverse function x=f^-1(y) we would calculate x knowing y.

The function can be written:

y=x+cos(x)

and since we know what y is we plug that in:

1=x+cos(x)

Now we dont know what the inverse function is and it doesnt look
easy to calculate or it doesnt even exist, so we look at the equation
above and try to solve it. We need to find an x that when used in
"x+cos(x)" it comes out to an answer of 1.
The only solution i can see is x=0, because this is true:
1=0+cos(0)

so that means that f^-1(1)=0 because when y equals one x equals zero.

#### j.friend

##### New Member
I'm sorry but that doesn't seem entirely correct. Whilst the inverse may be hard to find, the value of the inverse function cannot be found be subbing 1 into the normal function. Whilst the substituion method may not be able to be utilised, the graphical analysis method may.

I'm sure you know that the inverse of a function is the function reflected over the line y = x. Either you can make a rough sketch by hand or use a graphical calcultor (common in most schools nowadays for senior mathematics courses).

By sketching the graph of f(x) = x + cosx, it can be seen that it will have an inverese function due to the graph being 1:1 (meaning that each x point has only one corresponding y point and vice versa).

Another aspect of inverse functions is that essentially the x and y values swap for the inverse. This means that for the numbers subbed in (if y = 1, x = 0) on the inverse graph there will be a point (0,1) or at x = 0, y = 1.

Taking this into account by finding the value of the original function when x = 1, the value for the inverse function when y = 1 will be found.

so y = x + cosx
if x = 1, y = 1 + cos 1 (your calculator will need to be in radians mode to calculate this properly)

from this you get y ≈ 1.54

this means that the value of f^-1(1) ≈ 1.54

#### MrAl

##### Well-Known Member
Hi

Ummm, and how exactly are you utilizing the inverse function there?
I dont see any inverse function in your solution...did i miss something?

The original function is:

y=x+cos(x)

or, stated another way:

y=f(x)

The problem was to find the value of the *inverse* function when
y is made equal to 1, not to simply substitute 1 in for x. That is,
if we knew what f' was we could then (and only then) plug 1 into

#### j.friend

##### New Member
ah yes,

I misread/misinterpretted what was going on. Rather silly of me.

I apologise

#### Ariek

##### New Member
Mr. Al, thanks so much! I had same hw question, but how would I then find deriv of g(1)?

#### MrAl

##### Well-Known Member
Hi,

It looks like part of this thread is missing, the very beginning. It's also around 8 years old

#### Ratchit

##### Well-Known Member
Hi,

f(x)=x+cos(x)

and it looks like you need to find f^-1(1) which would mean if we
had the inverse function x=f^-1(y) we would calculate x knowing y.

The function can be written:

y=x+cos(x)

and since we know what y is we plug that in:

1=x+cos(x)

Now we dont know what the inverse function is and it doesnt look
easy to calculate or it doesnt even exist, so we look at the equation
above and try to solve it. We need to find an x that when used in
"x+cos(x)" it comes out to an answer of 1.
The only solution i can see is x=0, because this is true:
1=0+cos(0)

so that means that f^-1(1)=0 because when y equals one x equals zero.

Sometimes a plot helps. As you can see, there is only one solution to the inverse of the function.

Ratch

#### MrAl

##### Well-Known Member
Sometimes a plot helps. As you can see, there is only one solution to the inverse of the function.

View attachment 105241

Ratch

Hello,

Yes a plot helps, but in this case you did not say how it helps. For example, are you saying that there are an infinite number fo solutions or just one? In other words, you need to state what your final solution to this problem would be.
We both know what it is, but someone else might not know what you are implying with the graph.
BTW it's a little clearer to graph y=f(x)-y1 rather than y=f(x).

Back to main ideas...

The derivative of the inverse function can be found using the property that the inverse function slope is 1/m where m is the slope of the original function. Watch the sign, and probably you want the derivative at the solution point.

Last edited:

#### Ratchit

##### Well-Known Member
Hello,

Yes a plot helps, but in this case you did not say how it helps. For example, are you saying that there are an infinite number fo solutions or just one? In other words, you need to state what your final solution to this problem would be.
We both know what it is, but someone else might not know what you are implying with the graph.
BTW it's a little clearer to graph y=f(x)-y1 rather than y=f(x).

Back to main ideas...

The derivative of the inverse function can be found using the property that the inverse function slope is 1/m where m is the slope of the original function. Watch the sign, and probably you want the derivative at the solution point.

Where does it say that the derivative is involved anywhere? I thought the remark by the OP about f ' was a misinterpretation. If you think otherwise, would you please restate the problem more understandably? Until I see otherwise, I am assuming the problem is asking for a value when the the inverse of f(x), call it g(x), is g(1). As you observed, it is easily seen by inspection that g(1) = 0. Furthermore, the plot confirms that zero is the only solution to g(1), and shows there is a unique y-value for each x-value and vice-versa. I don't know what more can be said about this.

Ratch

#### MrAl

##### Well-Known Member
Hi Ratch,

In post #6 it was asked how to find the "deriv" of g.

What you say now is true, but before that you did not explain what the graph was doing for us. People who dont do this that much dont know what they are looking at. For example, in your graph do we look for the y axis crossing or the x axis crossing or consider all the points on that graph to mean something. Until you explain it someone might not know. Now that you've explained it it will be clear to almost anyone. You have to keep in mind you might be dealing with students who are new to this. I think it is very good to show the graph though, with a short added explanation.

There are some other interesting ways to solve this too, but not sure if i want to get into that right now.

#### Ratchit

##### Well-Known Member
Hi Ratch,

In post #6 it was asked how to find the "deriv" of g.

What you say now is true, but before that you did not explain what the graph was doing for us. People who dont do this that much dont know what they are looking at. For example, in your graph do we look for the y axis crossing or the x axis crossing or consider all the points on that graph to mean something. Until you explain it someone might not know. Now that you've explained it it will be clear to almost anyone. You have to keep in mind you might be dealing with students who are new to this. I think it is very good to show the graph though, with a short added explanation.

There are some other interesting ways to solve this too, but not sure if i want to get into that right now.

OK, the plot below is the function f(x).

The next plot in the inverse of f(x) which I will call g(x).

The last plot is the derivative of g'(x)

Regards,

Ratch

#### kalawi_tan

##### New Member
Woah! This is awesome! I missed my college days with Algebra, Physics, Advaced Mathematics, Thermodynamics, and such..

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