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


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