Increasing and Decreasing Intervals (Functions)

If the line of the graph starts below and, as it moves to the right it goes up, that means that the function is increasing. That is, the function grows when the values of $Y$ increase as those of $X$ grow (that is, move from left to right)

If the line of the graph starts at the top and, as it moves to the right it goes down, that means the function is decreasing. That is, the function decreases when the values of $Y$ go down as those of $X$ increase (that is, move from left to right)

If the line on the graph starts at a certain point on the $Y$ axis, and as it moves to the right it remains constant at the same height, that is, at the same point on the $Y$ axis, this means that it is a constant function. That is, the function is constant when the values of $Y$ keep their place and remain fixed as those of $X$ increase (that is, move from left to right)

To identify the intervals where the function is increasing, we will look on the graph for the point where the function begins to rise.

We will mark the value on the $X$ axis. In our case, it is $-5$. Then, we will look on the $X$ axis for the point where the function stops rising. In our case, it is $7$. Therefore, the growth interval of the function will be: 

$-5<X<7$

We will illustrate this with a simple graph: 

In the graph, it can be seen that the intervals of growth of the function are $X<-3$ (values of $X$ less than $-3$) and for the values of $X$ that are between $0$ and $3$. That is, in these intervals, the values of $X$ and $Y$ increase together. 

Furthermore, it follows from the graph that the intervals of decline of the function are for the values of $X$ that are between $-3$ and $0$ and for $X>3$. That is, in these intervals, the values of $X$ increase and those of $Y$ decrease at the same time.  

Exercise

Note that, in the graph, you can also see the intervals of decline of the function. Do you know what they are?

Answer

$-10<X<-5$

$7<X<10$

To identify the intervals where the function is decreasing, we will look on the graph for the point where the function starts to go down.

We will mark the value on the $X$ axis. In our case, it is $7$. Then we will look on the $X$ axis for the point where the function stops going down. In our case, it is $5$. Therefore, the interval of decrease of the function will be:

$-7<X<5$

Exercise

Notice that, on the graph, you can also see the intervals of increase of the function. Do you know what they are?

Answer

$-10<X<-7$

$5<X<10$

If you are interested in this article, you might also be interested in the following articles:

Graphical representation of a function

Algebraic representation of a function

Notation of a function

Domain of a function

Indefinite integral

Assignment of numerical value in a function

Variation of a function

Increasing function

Decreasing function

Constant function

Functions for seventh grade

Intervals of increase and decrease of a function

In the blog of Tutorela you will find a variety of articles with interesting explanations about mathematics

Assignment

Find the increasing area of the function

$f(x)=6x^2-12$

Solution

In the first step, let's consider that $a=6$

Therefore $a>0$ and the parabola is at a minimum

In the second step, we find $x$ of the vertex

according to the data we know that:

$a=6,b=0,c=-12$

We replace the data in the formula

$x=\frac{-b}{2\cdot a}$

$x=\frac{-0}{2\cdot6}$

$x=\frac{0}{12}$

$x=0$

Therefore

$0<x$ Increasing

$x<0$ Decreasing

Answer

$0<x$

Assignment

Given the function in the diagram, what is its domain of positivity?

Solution

Note that the entire function is always above the axis: $x$

Therefore, it will always be positive. Its area of positivity will be for all $x$

Answer

For all $x$

Assignment

Find the increasing area of the function

$f(x)=-4x^2-24$

Solution

In the first step, let's consider that $a=-4$

Therefore $a<0$ and the parabola is at its maximum

In the second step, we find $x$ of the vertex

according to the data we know that:

$a=-4,b=0,c=-24$

We replace the data in the formula

$x=\frac{-b}{2\cdot a}$

$x=\frac{-0}{2\cdot\left(-4\right)}$

$x=\frac{0}{-8}$

$x=0$

Therefore $x<0$ increasing area

Answer

$x<0$

Assignment

Find the increasing area of the function

$f(x)=2x^2$

Solution

In the first step, let's consider that $a=2$

Therefore $a>0$ and the parabola is minimum

In the second step, we find $x$ of the vertex

according to the data we know that:

$a=2,b=0,c=0$

We replace the data in the formula:

$x=\frac{-b}{2\cdot a}$

$x=\frac{0}{2\cdot2}$

$x=\frac{0}{4}$

$x=0$

Therefore, there is increase in the area $0<x$

Answer

$0<x$

Assignment

Find the increasing area of the function

$f(x)=-3x^2+12$

Solution

In the first step, let's consider that $a=-3$

Therefore $a<0$ and the parabola is at its maximum

In the second step, we find $x$ of the vertex

according to the data we know that:

$a=3,b=0,c=12$

We replace the data in the formula

$x=\frac{-b}{2\cdot a}$

$x=\frac{-0}{2\cdot\left(-3\right)}$

$x=\frac{0}{-6}$

$x=0$

Therefore, there is increase in the area $x<0$

Answer

$x<0$

Assignment

Find the decreasing area of the function

$y=(x+1)+1$

Solution

$a$ coefficient of $x^2$

Therefore $0<a$

is the minimum point

The vertex of the function is $\left(-1,1\right)$

The function decreases in the area of $x<-1$

Answer

$x<-1$

Assignment

Given the function in the graph

When is the function positive?

Solution

The intersection point with the axis :$x$ is: $\left(-4,0\right)$

Positive before, then negative.

Therefore $x<-4$

Answer

$x<-4$