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