What kind of function is f (x) = 5?

Updated on : January 21, 2022 by Bella Johnston

What kind of function is f (x) = 5?

The given function f (x) = 5 is a constant function, that is, for each value of x, f (x) is always 5 (constant). If you graph it, you will get a straight line parallel to the x-axis at a distance of +5 units from the origin.

A constant function

Constant function: In mathematics, a constant function is a function whose value is the same for each input value.
Linear equations: The special type of linear function is the constant function ... it is a horizontal line.

This is a horizontal line. It has a slope of 0. Remember that f (x) is another way to describe y in ordered pairs. It is called constant and when the value of y is represented graphically it does not change, although the value of x will.

A continuous function is completely defined at a single point.

If you ask "what is math and / math for math x = whatever / math" you should get a single number.

Here, we ask if the above two functions are continuous at math x = 0 / math. As you should be able to guess from the hint I gave above, this means that in math x = 0 / math, both math e ^ x / math and math e ^ {- x} / math should produce the same value of math and / math. Let's check!

• math e ^ x / math in math x = 0 / math is math e ^ 0 / math
• mathematics e ^ {- x} / mathematics in mathematics x = 0 / mathematics is mathematics e ^ {- 0} = e ^ 0 / mathematics

Check it out! You get the same answer for both! math e ^ 0 / math (also known as math 1 / math) is the common y value for math e ^ x / math and math e ^ {- x} / math for math x = 0 / math.

The bottom line is, yes, these two functions are in fact continuous at math x = 0 / math.

Side note: if we

A continuous function is completely defined at a single point.

If you ask "what is math and / math for math x = whatever / math" you should get a single number.

Here, we ask if the above two functions are continuous at math x = 0 / math. As you should be able to guess from the hint I gave above, this means that in math x = 0 / math, both math e ^ x / math and math e ^ {- x} / math should produce the same value of math and / math. Let's check!

• math e ^ x / math in math x = 0 / math is math e ^ 0 / math
• mathematics e ^ {- x} / mathematics in mathematics x = 0 / mathematics is mathematics e ^ {- 0} = e ^ 0 / mathematics

Check it out! You get the same answer for both! math e ^ 0 / math (also known as math 1 / math) is the common y value for math e ^ x / math and math e ^ {- x} / math for math x = 0 / math.

The bottom line is, yes, these two functions are in fact continuous at math x = 0 / math.

Side note: If you were asked whether or not these two functions are differentiable in math x = 0 / math, their respective derivatives in math x = 0 / math should be the same.

Edit: I just realized that this answer assumes piecemeal behavior and doesn't actually answer the question.

The real answer is, math e ^ x / math is continuous at math x = 0 / math because there is no reason why it should be discontinuous. A function is discontinuous when it is not completely defined at a point. But math e ^ x / math is fully defined in math x = 0 / math: we know that it is math 1 / math. The same goes for math e ^ {- x} / math.

If there are no solutions or if there are several unique solutions for math and / math in math x = 0 / math, then you have discontinuity. Here we have a single solution, since we would all expect a typical function to behave.

There is no "the" here. There are infinite math functions f (x) / math such that math f (f (x)) = e ^ x / math for all math x \ in \ R / math, but none of them are particularly nice or have a simple or elementary expression. They are sometimes called "semi-exponential functions," and they have earned celebrity status by having their own Wikipedia article.

Here, let's build one of those functions.

We want math f (f (0)) = 1 / math, so let's make the natural choice math f (0) = \ frac {1} {2} / math and math f \ left (\ frac {1} {2} \ right) = 1 / math. What should we do with numbers between math 0 / math and math \ frac {1} {2} / math? Why, we can do whatever we want as long as we keep them out of that range, so that we can

There is no "the" here. There are infinite math functions f (x) / math such that math f (f (x)) = e ^ x / math for all math x \ in \ R / math, but none of them are particularly nice or have a simple or elementary expression. They are sometimes called "semi-exponential functions," and they have earned celebrity status by having their own Wikipedia article.

Here, let's build one of those functions.

We want math f (f (0)) = 1 / math, so let's make the natural choice math f (0) = \ frac {1} {2} / math and math f \ left (\ frac {1} {2} \ right) = 1 / math. What should we do with numbers between math 0 / math and math \ frac {1} {2} / math? Why, we can do whatever we want as long as we move them away from that range, so that we can choose the appropriate iterated values. So let's say

math f (x) = x + \ frac {1} {2} / math when math 0 \ le x \ le \ frac {1} {2} / math.

Well. All these numbers move to the interval mathematics 1 / 2,1 / mathematics, so in that interval our hand is forced: we must choose

math f \ left (x + \ frac {1} {2} \ right) = e ^ x / math, math 0 \ le x \ le \ frac {1} {2} / math

or in other words

math f (x) = e ^ {x-1/2} / math when math \ frac {1} {2} \ le x \ le 1 / math.

This, in turn, returns the numbers in math 1 / 2,1 / math in the interval math 1, e ^ {1/2 / math, so a once again our hand is forced in that interval:

math f \ left (e ^ {x-1/2} \ right) = e ^ x / math, math \ frac {1} {2} \ le x \ le 1 / math

if we call math u = e ^ {x-1/2} / math then math x = \ ln (u) +1/2 / math, then math f (u) = \ exp (\ ln (u) +1/2) = u \ sqrt {e} / math. Rewriting this with math x / math as a variable,

math f (x) = \ sqrt {e} x / math when math 1 \ le x \ le \ sqrt {e} / math

So far our function looks like this:

The red and green segments are straight lines, while the blue is a constant factor multiplied by exponential.

We can keep repeating the same procedure. There is no succinct way to write the entire function unless we resort to a recursive expression:

math f (x) = \ exp (f (\ ln x)) / math when math 1 \ le x / math

This is not circular because in any given interval, math \ ln x / math is in the previous interval. Similary,

math f (x) = \ ln (f (\ exp x)) / math when math x \ le 0 / math

In general, this is a monotonically increasing, continuous exponential half, but it is not smooth and does not have a simple elementary expression.

Basically it really depends on what domain you are in.

Here is a brief construction for the domain of N natural numbers.
I hope it helps.

We denote "f (a) = b" as "a -> b" below.

1.
f (f (0)) = 0: 0 -> 0

2.
To have f (f (1)) = 2, there
must be 1 -> T1 -> 2,

To have f (f (2)) = 6, there
must be 2 -> T2 -> 6

Keep doing it, we have a mapping chain:
1 -> T1 -> 2 -> T2 -> 6 -> T3 -> 42 -> T4 -> ...

We note that there are many series of this type:
1, 2, 6, 42, ...
3, 12, 156, 24492, ...
4, 20, 420, ...
......

They all come from continuing to apply "x ^ 2 + x".
They are all disjointed and the union is Z +.

3.
We can inject any "

Basically it really depends on what domain you are in.

Here is a brief construction for the domain of N natural numbers.
I hope it helps.

We denote "f (a) = b" as "a -> b" below.

1.
f (f (0)) = 0: 0 -> 0

2.
To have f (f (1)) = 2, there
must be 1 -> T1 -> 2,

To have f (f (2)) = 6, there
must be 2 -> T2 -> 6

Keep doing it, we have a mapping chain:
1 -> T1 -> 2 -> T2 -> 6 -> T3 -> 42 -> T4 -> ...

We note that there are many series of this type:
1, 2, 6, 42, ...
3, 12, 156, 24492, ...
4, 20, 420, ...
......

They all come from continuing to apply "x ^ 2 + x".
They are all disjointed and the union is Z +.

3.
We can inject any "unused" series into "T1, T2, T3, ..." in the above mapping chain.

For example, we inject
3, 12, 156, 24492, ...
into
1 -> T1 -> 2 -> T2 -> 6 -> T3 -> 42 -> T4 -> ...

We have:
1 -> 3 -> 2 -> 12 -> 6 -> 156 -> 42 -> 24492 -> ...

For now we have built a mapping chain using two series "1, 2, 6, 42, ..." and "3, 12, 156, 24492, ...".

4.
We continue to choose two unused series and combine them into a new mapping chain.

By continuing to do this, we inductively construct f (x) in Z +.

Finally,
obviously the solution is not unique.
And I think this conclusion also applies to the real numbers R and there is a similar construction.

A relation is a set of pairs ordered so that y = 5 has the following set of solutions {(1,5), (2,5), (66,5), (-42, 5), (pi, 5 ), (SR2, 5) …… without any domain elements "going" to two or more range elements}. Etc.

Note that y = 5 constitutes a set of ordered pairs. Now, a relation is a function if each element in the domain is "associated" with exactly one element in the range. Well, the above relation is a function since for example 66 is associated with 5 and not another. The same for (2,5) etc.

To see a relation that is not a function, look at {(2,6), (5, 7), (2,99)… ..}. Note that domain element 2 is

A relation is a set of pairs ordered so that y = 5 has the following set of solutions {(1,5), (2,5), (66,5), (-42, 5), (pi, 5 ), (SR2, 5) …… without any domain elements "going" to two or more range elements}. Etc.

Note that y = 5 constitutes a set of ordered pairs. Now, a relation is a function if each element in the domain is "associated" with exactly one element in the range. Well, the above relation is a function since for example 66 is associated with 5 and not another. The same for (2,5) etc.

To see a relation that is not a function, look at {(2,6), (5, 7), (2,99)… ..}. Note that domain element 2 is "paired / associated" with both 6 and 99, which does not follow the definition of a function.

y = x ^ 2 is a relation that is a function, but….

x = y ^ 2 is a relation that is NOT a function

Try to find some ordered pairs of both relations above and convince yourself that only one of the two is a function

Just to make it a little clearer (we'll have less x running), let's say the function is f (w) = w - 1 / w.

You are asking why f (1 / x) = -f (x).

When you plug in 1 / x for w, what do you get?

1 / x - 1 / (1 / x).

That is 1 / x - x

When you connect x by w, you get x - 1 / x.

So -f (x) = - (x - 1 / x), and you have to show that that is equal to 1 / x - x.

Consider f (x): = a * x for a # 0 (since that reduces to f (x): = 0). Then f (5 * x) = a * (5 * x) = 5 * (a * x) =

5 * f (x). So any linear function of x, even for a = 0, works with f (x): = a * x satisfying the condition f (5 * x) = 5 * f (x).

Although search engines are notoriously bad at looking for mathematical expressions, if you scroll through the pages, the result you are looking for is often hidden.

How do you solve \$ f '(x) = f (f (x)) \$?

What is a function of f (x) so that the curvature (f (x)) = f (X)?

The function f (x) = 0, x∈ℝ meets your requirements.

1 / x - Wolfram | Alpha

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