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Let *I* be an open interval contained in the domain of the real valued function y= f(x). Then:

*i) * *f* is said to be increasing on I if x_{1} < x_{2} in ⇒ f(x_{1}) ≤ f(x_{2}) ∀ x_{1},x_{2} ∈ I

*ii) * *f* is said to be a strictly increasing function on I if x_{1} < x_{2} ⇒ f(x_{1}) < f(x_{2}) ∀ x_{1},x_{2} ∈ I

*iii) * f is said to be a decreasing function on I if x_{1} < x_{2} ⇒ f(x_{1}) ≥ f(x_{2}) ∀ x_{1},x_{2} ∈ I

*iv) * f is said to be a strictly decreasing function on I if x_{1} < x_{2} ⇒ f(x_{1}) > f(x_{2}) ∀ x_{1},x_{2} ∈ I

Take a graph f(x) = sin x

From the graph between 0 and π you will see that as *x increases* from 0 to π/2, sin x increases from 0 to one.

That is, the height of the graph increases as we move from 0 to π/2 on the x-axis.

And, as *x* increases from π/2 to π, sine *x* decreases from 1 to 0.

In other words, the height of the graph decreases as we move from π/2 to π on the x-axis.

Ex:

Is f(x) = sin x an increasing function or a decreasing function?

**Sol** :

Domain of sin x = R.

Range = [-1, 1]

Let I_{1} = (0 , π/2) ∈ R and I_{2} = (π/2 , π) ∈ R

Let x_{1} = π/4 and x_{2 }= π/3 ∈ I_{1}

⇒ f(x_{1}) = 1/√2 and f(x_{2}) = √3/2

Then f(x_{1}) < f(x_{2}) ∀ x_{1,}x_{2} ∈ (0,π/2)

Hence sin x is strictly increasing on interval I.

Now we consider two points x_{1 = }2π/3 and x_{2} = 3π/4 ∈ I_{2}

⇒ f(x_{1}) = √3/2 and f(x_{2}) = 1/√2

Then f(x_{1}) > f(x_{2}) ∀ x_{1,}x_{2} ∈ (π/2, π)

Hence sin x is strictly decreasing on interval I.

**Note**: sin x is neither increasing nor decreasing on R.

The graph of an increasing function is as shown.

The straight line in the graph depicts that at some points That is, x_{1 < }x_{2} in ⇒ f(x_{1}) = f(x_{2}).

Hence, this function is not strictly increasing.

Example :

Consider f(x) = x + 2

x |
-1 |
0 |
1 |
2 |

f(x) |
1 |
2 |
3 |
4 |

For every value of x, we get a different value for f(x).

And, as the value of x is increasing from -1 to 2, the value of f(x) is increasing from one to four.

So, for every x_{1} < x_{2} ⇒ f(x_{1}) < f(x_{2}) ∀ x_{1},x_{2} ∈ I

Hence, f(x) = x + 2 is strictly increasing on the set of real numbers.

Now, when a function is said to be decreasing or strictly decreasing on an interval.

The graph of a decreasing function.

The straight line on the graph depicts that with the increase in the value of x, the value of the

function remains constant at some points.

Consider the function f(x) = -x + 2.

Observe that as the value of x increases from -1 to 2, the value of f(x) decreases from 3 to 0.

So that value of x increases the value of the function decreases.

Then f(x_{1}) > f(x_{2}) ∀ x_{1},x_{2} ∈ R

Hence, f(x) = -x + 2 is a strictly decreasing function on the set of real numbers.

Let *I* be an open interval contained in the domain of the real valued function y= f(x). Then:

*i) * *f* is said to be increasing on I if x_{1} < x_{2} in ⇒ f(x_{1}) ≤ f(x_{2}) ∀ x_{1},x_{2} ∈ I

*ii) * *f* is said to be a strictly increasing function on I if x_{1} < x_{2} ⇒ f(x_{1}) < f(x_{2}) ∀ x_{1},x_{2} ∈ I

*iii) * f is said to be a decreasing function on I if x_{1} < x_{2} ⇒ f(x_{1}) ≥ f(x_{2}) ∀ x_{1},x_{2} ∈ I

*iv) * f is said to be a strictly decreasing function on I if x_{1} < x_{2} ⇒ f(x_{1}) > f(x_{2}) ∀ x_{1},x_{2} ∈ I

Take a graph f(x) = sin x

From the graph between 0 and π you will see that as *x increases* from 0 to π/2, sin x increases from 0 to one.

That is, the height of the graph increases as we move from 0 to π/2 on the x-axis.

And, as *x* increases from π/2 to π, sine *x* decreases from 1 to 0.

In other words, the height of the graph decreases as we move from π/2 to π on the x-axis.

Ex:

Is f(x) = sin x an increasing function or a decreasing function?

**Sol** :

Domain of sin x = R.

Range = [-1, 1]

Let I_{1} = (0 , π/2) ∈ R and I_{2} = (π/2 , π) ∈ R

Let x_{1} = π/4 and x_{2 }= π/3 ∈ I_{1}

⇒ f(x_{1}) = 1/√2 and f(x_{2}) = √3/2

Then f(x_{1}) < f(x_{2}) ∀ x_{1,}x_{2} ∈ (0,π/2)

Hence sin x is strictly increasing on interval I.

Now we consider two points x_{1 = }2π/3 and x_{2} = 3π/4 ∈ I_{2}

⇒ f(x_{1}) = √3/2 and f(x_{2}) = 1/√2

Then f(x_{1}) > f(x_{2}) ∀ x_{1,}x_{2} ∈ (π/2, π)

Hence sin x is strictly decreasing on interval I.

**Note**: sin x is neither increasing nor decreasing on R.

The graph of an increasing function is as shown.

The straight line in the graph depicts that at some points That is, x_{1 < }x_{2} in ⇒ f(x_{1}) = f(x_{2}).

Hence, this function is not strictly increasing.

Example :

Consider f(x) = x + 2

x |
-1 |
0 |
1 |
2 |

f(x) |
1 |
2 |
3 |
4 |

For every value of x, we get a different value for f(x).

And, as the value of x is increasing from -1 to 2, the value of f(x) is increasing from one to four.

So, for every x_{1} < x_{2} ⇒ f(x_{1}) < f(x_{2}) ∀ x_{1},x_{2} ∈ I

Hence, f(x) = x + 2 is strictly increasing on the set of real numbers.

Now, when a function is said to be decreasing or strictly decreasing on an interval.

The graph of a decreasing function.

The straight line on the graph depicts that with the increase in the value of x, the value of the

function remains constant at some points.

Consider the function f(x) = -x + 2.

Observe that as the value of x increases from -1 to 2, the value of f(x) decreases from 3 to 0.

So that value of x increases the value of the function decreases.

Then f(x_{1}) > f(x_{2}) ∀ x_{1},x_{2} ∈ R

Hence, f(x) = -x + 2 is a strictly decreasing function on the set of real numbers.