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 $f(x_1) < f (x_2) \; \forall x_1 , x_2 \in I$.

(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) $

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 $f(x_1) > f (x_2) \forall x_1 , x_2 \in I$.

(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 $f(x_1) > f (x_2)$

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 $x \in (a, b)$

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

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 $x \in (a, b)$

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

- 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.

Determine the intervals over which the function \(f(x) = x^3 - 3x^2 - 4x + 12\) is strictly increasing or decreasing.

\(f(x) = x^3 - 3x^2 - 4x + 12\)

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 \(-\frac{1}{3} < x < 4\).

Find the intervals over which the function \(f(x) = \sin x + \cos x\) is increasing or decreasing in \([0, 2\pi]\).

\(f(x) = \sin x + \cos x\)

\[f'(x) = \cos x - \sin x\]

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 \([\frac{\pi}{4}, \frac{5\pi}{4}]\).

Find the intervals of increase and decrease for the function \(f(x) = e^{-x}(x^2 - 2x)\).

\(f(x) = e^{-x}(x^2 - 2x)\).

\[f'(x) = e^{-x}(x^2 - 4x + 2)\]

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 \(1 - \sqrt{3} < x < 1 + \sqrt{3}\).

**Notes**-
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