# Increasing and Decreasing functions

## Increasing Functions

Let I be an interval contained in the domain of a real valued function $f$.
(i) Then f is said to be increasing on I
if $x_1 < x_2$ in I
and $f(x_1) \leq f (x_2) \; \forall x_1 , x_2 \in I$.
(ii) Then f is said to be strictly increasing on I
if $x_1 < x_2$ in I
and $f(x_1) < f (x_2) \; \forall x_1 , x_2 \in I$.

## Increasing Functions at a point

Let $x_0$ be a point in the domain of definition of a real valued function $f$.
(i) Then the function $f$ is said to be increasing at $x_0$
If there exists an interval $I = (x_0 - h, x_0 + h)$, h > 0 such that $\forall x_1 , x_2 \in I$.
$x_1 < x_2$ and $f(x_1) \leq f (x_2)$
(ii)Then the function $f$ is said to be strictly increasing at $x_0$
If there exists an interval $I = (x_0 - h, x_0+ h)$, h > 0 such that $\forall x_1 , x_2 \in I$.
$x_1 < x_2$ and $f(x_1) < f (x_2)$

## Decreasing Functions

Let I be an interval contained in the domain of a real valued function $f$.
(i) Then f is said to be decreasing on I
if $x_1 < x_2$ in I
and $f(x_1) \geq f (x_2) \forall x_1 , x_2 \in I$.
(ii) Then f is said to be strictly decreasing on I
if $x_1 < x_2$ in I
and $f(x_1) > f (x_2) \forall x_1 , x_2 \in I$.

## Decreasing Functions at a point

Let $x_0$ be a point in the domain of definition of a real valued function f.
(i) Then the function $f$ is said to be Decreasing at $x_0$
If there exists an interval $I = (x_0 - h, x_0 + h)$, h > 0 such that $\forall x_1 , x_2 \in I$.
$x_1 < x_2$ and $f(x_1) \geq f (x_2)$
(ii)Then the function $f$ is said to be strictly decreasing at $x_0$
If there exists an interval $I = (x_0 - h, x_0+ h)$, h > 0 such that $\forall x_1 , x_2 \in I$.
$x_1 < x_2$ and $f(x_1) > f (x_2)$

## Increasing function using First derivative

Theorem
Let $f$ be continuous on [a, b] and differentiable on the open interval (a,b). Then f is strictly increasing in [a,b]
if f′(x) > 0 for each $x \in (a, b)$
Proof
Let $x_1, x_2 \in [a, b]$ be such that $x_1 < x_2$
Then, by Mean Value Theorem there exists a point c between $x_1$ and $x_2$ such that
$f'(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}.$
So
$f(x_2) - f(x_1)= f'c(x_2 -x_1)$
If f'(c) > 0
Then
$f(x_2) - f(x_1) > 0$
$f(x_2) > f(x_1)$
So Strictly Increasing.
Hence Proved

## Decreasing function using First derivative

Theorem
Let $f$ be continuous on [a, b] and differentiable on the open interval (a,b). Then f is strictly decreasing in [a,b] if f′(x) < 0 for each $x \in (a, b)$ Proof
Let $x_1, x_2 \in [a, b]$ be such that $x_1 < x_2$
Then, by Mean Value Theorem there exists a point c between $x_1$ and $x_2$ such that
$f'(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}.$
So
$f(x_2) - f(x_1)= f'c(x_2 -x_1)$
If f'(c) < 0
Then
$f(x_2) - f(x_1) < 0$
$f(x_2) < f(x_1)$
So Strictly Decreasing.
Hence Proved

## Increasing and Decreasing Functions

• A function can be increasing and decreasing in various interval.
• By analyzing the sign of the first derivative, we can determine intervals where the function is increasing or decreasing.
• If f' (x)< 0 in an interval, then f(x) is increasing in that interval.
• If f' (x) <0 in an interval, then f(x) is decreasing in that interval.

## Solved Examples

Question 1
Determine the intervals over which the function $f(x) = x^3 - 3x^2 - 4x + 12$ is strictly increasing or decreasing.
Solution
$f(x) = x^3 - 3x^2 - 4x + 12$
So,
$f'(x) = 3x^2 - 6x - 4$
Lets Determine the values of $x$ for which $f'(x) = 0$.
$3x^2 - 6x - 4 = 0$
$x = -\frac{1}{3} \text{ and } x = 4$
Now checking increasing and decreasing Using test points:
For $x < -\frac{1}{3}$, $f'(x) > 0$. Function is increasing.
For $-\frac{1}{3} < x < 4$, $f'(x) < 0$. Function is decreasing.
For $x > 4$, $f'(x) > 0$. Function is increasing.
Hence
The function is increasing for $x < -\frac{1}{3}$ and $x > 4$ and decreasing for $-\frac{1}{3} < x < 4$.

Question 2
Find the intervals over which the function $f(x) = \sin x + \cos x$ is increasing or decreasing in $[0, 2\pi]$.
Solution
$f(x) = \sin x + \cos x$
$f'(x) = \cos x - \sin x$
Lets Determine the values of $x$ for which $f'(x) = 0$.
Solving $\cos x = \sin x$, we get $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.
Now checking increasing and decreasing Using test points:
For $0 < x < \frac{\pi}{4}$, $f'(x) > 0$. Increasing.
For $\frac{\pi}{4} < x < \frac{5\pi}{4}$, $f'(x) < 0$. Decreasing.
For $\frac{5\pi}{4} < x < 2\pi$, $f'(x) > 0$. Increasing.
Hence The function is increasing in the intervals $[0, \frac{\pi}{4}]$ and $[\frac{5\pi}{4}, 2\pi]$ and decreasing in $[\frac{\pi}{4}, \frac{5\pi}{4}]$.

Question 2
Find the intervals of increase and decrease for the function $f(x) = e^{-x}(x^2 - 2x)$.
Solution
$f(x) = e^{-x}(x^2 - 2x)$.
$f'(x) = e^{-x}(x^2 - 4x + 2)$
Lets Determine the values of $x$ for which $f'(x) = 0$.
$x^2 - 4x + 2 = 0$
This gives $x = 1 + \sqrt{3}$ and $x = 1 - \sqrt{3}$.
Now checking increasing and decreasing Using test points:
For $x < 1 - \sqrt{3}$, $f'(x) > 0$. Increasing.
For $1 - \sqrt{3} < x < 1 + \sqrt{3}$, $f'(x) < 0$. Decreasing.
For $x > 1 + \sqrt{3}$, $f'(x) > 0$. Increasing.
Therefore
The function increases for $x < 1 - \sqrt{3}$ and $x > 1 + \sqrt{3}$, and decreases for $1 - \sqrt{3} < x < 1 + \sqrt{3}$.

## Related Topics

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