﻿ Image of a function. What it is and how it is calculated. Exercises resolved.

# Image of a function. What it is and how to calculate it step by step. Exercises resolved.

What is the image of a function and how is the image of a function calculated? I will explain step by step what the image of a function is and show you how to calculate it.

## What is the image of a function

In order to understand very well what the image of a function is, you must first be very clear about what the mastery of a function is.

I remind you that the domain of a function is the range of values of x for which f(x) exists, i.e. the values of x, for which f(x) has a result.

Graphically, the domain is looked at on the x-axis, as these are the x-values for which the function exists, i.e. the function is represented above.

For example, we have the following function: What will be your domain name?

Your domain is the range of x values for which the function is represented above (the function exists). I’ll mark it in blue: Therefore, the mastery of that function is: Now, what is the image of this function? And by the way, what is the image of a performance?

The image is the range of values of f(x) for which there is a value of x[/box].

It is designated as Im f: The image of a function can also be called a path or range.

In other words, they are the values of f(x) in which the function exists.

Graphically you look at the y-axis, (since f(x) and y, it’s the same). Therefore, the function image of the above example is the values that are on the y-axis, for which the function exists. I’ll mark it in green: So the image of that function would be:

##  How is the image of a function calculated?

The image is closely related to the domain of a function, since to calculate the image, it is necessary to calculate the domain beforehand.

The way it is calculated will depend on the type of function. Let’s go see him:

### Image of polynomial functions

In these functions, the domain is all R, so the image is also all R: f(x) always exists, its domain is all R, so the image is: As long as the domain is all R, the image will be all R, except for the quadratic (second degree) functions, which I explain at the end of the lesson.

### Irrational functions image

In irrational functions such as, for example: To calculate the image we must first calculate your domain.

The mastery of this function is: Once we have the domain, we calculate the value of f(x) for each end of the domain: These two values of f(x), correspond to the ends of the range of values in the image, so the image of that function is: By the way, this example of an irrational function corresponds to the graph in the previous example. You can check how the calculated domain and the image match the ones we obtained graphically.

### Image of logarithmic functions

The picture of logarithmic functions is all R, by definition, regardless of your domain.

### Image of rational functions

Calculating the image of rational functions is somewhat more complex than the previous cases. To calculate the domain of a rational function, we must first calculate its inverse function, which is designated as: Between a function and its inverse, these two conditions are met:

1- The domain of the inverse function is equal to the image of f: 2- The image of f-1 is the domain of f: Therefore, once we have obtained the inverse function, we have to calculate its domain, since the domain of the inverse function will be the image of the original function.

Here is a step-by-step example: What is the image of the next function? Let’s calculate its inverse function. The domain of the inverse function will be the image of this function.

Steps to calculate the inverse of a function:

We call f(x) y: Now the x must be cleared, i.e. it must be left alone in the first member. To do this, we pass the 1-x denominator by multiplying it to the first member and the 1-x denominator by dividing it into the second member: Now we leave the x alone, passing the 1 minus the second member: Once we have the x cleared, we exchange the unknowns: to the x we call y, and to the y we call x: And this is the reverse function. Therefore a and we call it f elevated to -1: Now that we have the reverse function, we can calculate your domain: The domain of the inverse function is the image of the original function: ## Image of quadratic functions or second degree functions.

To calculate the image of the second degree functions, you could also calculate it by obtaining the mastery of its inverse function, but we will calculate it with another method.

The graph of a quadratic function is a parabola. So:

• If its vertex is pointing downwards, its vertex will mark the lowest point on the y-axis
• If its vertex is pointing upwards, its vertex will mark the maximum point on the y-axis

Let’s look at it with an example. Calculate the image of the next function: In a quadratic function of the form: The value of x where the vertex is located is calculated with the following formula: In our function, the value of x where the vertex is located is: Therefore, the vertex will be at x=-1

What’s the function worth at that point?

We replace the x with -1 in the function: At x=-1, the function (the value of y) is 0. The vertex is therefore at point (-1.0).

Is 0 the maximum or minimum value of the function?

Let’s give any value to x, to see if the function takes a value bigger or smaller than 0. We give for example the value of x=1: The function gives 4, which is greater than 0, so 0 is the minimum value.

So as the image is looked at on the y-axis, and 0 is the smallest value the function can be worth, the image will be: I’ll leave you the graph of the show to make it clearer: The same thing happens with bi-quadratic functions such as for example: And in general in functions where the grade of the first term is twice as high as the grade of the second term as: When the quadratic functions are not complete, we can also calculate the image from its inverse function. For example: We calculate its inverse function: We found the domain of the reverse function: Therefore, the image of the original function is: 