Increasing and Decreasing Functions

Definitions of Monotonic Functions

A function is said to be monotonic if it is either entirely non-increasing or non-decreasing on a given interval. Let $I$ be an interval contained in the domain of a real-valued function $f$. Then $f$ is said to be:

• Strictly Increasing: if $x_1 < x_2$ in $I$ implies $f(x_1) < f(x_2)$ for all $x_1, x_2 \in I$.
• Increasing (Non-decreasing): if $x_1 < x_2$ in $I$ implies $f(x_1) \le f(x_2)$ for all $x_1, x_2 \in I$.
• Strictly Decreasing: if $x_1 < x_2$ in $I$ implies $f(x_1) > f(x_2)$ for all $x_1, x_2 \in I$.
• Decreasing (Non-increasing): if $x_1 < x_2$ in $I$ implies $f(x_1) \ge f(x_2)$ for all $x_1, x_2 \in I$.
• Constant: if $f(x) = c$ for all $x \in I$, where $c$ is a constant.

These definitions are purely algebraic and do not require calculus. Derivatives are then used as a powerful test to identify such behaviour.

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First Derivative Test for Monotonicity

Calculus provides a very powerful tool to determine the intervals on which a function is increasing or decreasing using its first derivative.

Theorem: Let $f$ be continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$. Then:

1. $f$ is strictly increasing on $[a, b]$ if $f'(x) > 0$ for each $x \in (a, b)$.
2. $f$ is strictly decreasing on $[a, b]$ if $f'(x) < 0$ for each $x \in (a, b)$.
3. $f$ is a constant function on $[a, b]$ if $f'(x) = 0$ for each $x \in (a, b)$.

Useful extension:

• If $f'(x) \ge 0$ for all $x \in (a, b)$, then $f$ is increasing (non-decreasing) on $[a, b]$.
• If $f'(x) \le 0$ for all $x \in (a, b)$, then $f$ is decreasing (non-increasing) on $[a, b]$.

So, the sign of $f'(x)$ tells us the sign of the slope of the tangent, and hence whether the function rises or falls.

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Working Rule for Finding Intervals

To find the intervals in which a function $f(x)$ is strictly increasing or strictly decreasing, follow these steps:

Step 1: Find the derivative $f'(x)$ of the given function $f(x)$.

Step 2: Set $f'(x) = 0$ and solve for $x$ to find the critical points. Also note any points where $f'(x)$ is undefined but $f(x)$ exists.

Step 3: These critical points divide the domain into disjoint intervals.

Step 4: Select a test point in each interval and evaluate the sign of $f'(x)$.

• If $f'(x) > 0$ in the interval, then $f(x)$ is strictly increasing there.
• If $f'(x) < 0$ in the interval, then $f(x)$ is strictly decreasing there.

Instead of always substituting numerical test points, you may also inspect the sign directly from the factorized form of $f'(x)$ whenever convenient.

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Solved Examples

Example 1: Finding Intervals for a Quadratic Function

Find the intervals in which the function $f$ given by $f(x) = x^2 - 4x + 6$ is (a) strictly increasing (b) strictly decreasing.

Solution: Step 1: Find the first derivative $f'(x)$. $$f'(x) = \frac{d}{dx}(x^2 - 4x + 6) = 2x - 4.$$

Step 2: Find the critical points by setting $f'(x) = 0$. $$2x - 4 = 0 \implies 2x = 4 \implies x = 2.$$

Step 3: The point $x = 2$ divides the real line into two disjoint intervals: $(-\infty, 2)$ and $(2, \infty)$.

Step 4: Test the sign of $f'(x)$ in each interval.

• Interval $(-\infty, 2)$: Pick a test point $x = 0$. $$f'(0) = 2(0) - 4 = -4 < 0.$$ Hence, $f(x)$ is strictly decreasing in $(-\infty, 2)$.
• Interval $(2, \infty)$: Pick a test point $x = 3$. $$f'(3) = 2(3) - 4 = 2 > 0.$$ Hence, $f(x)$ is strictly increasing in $(2, \infty)$.

Answer: (a) Strictly increasing in $(2, \infty)$. (b) Strictly decreasing in $(-\infty, 2)$.

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Example 2: Finding Intervals for a Cubic Function

Find the intervals in which the function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly increasing or strictly decreasing.

Solution: Step 1: Find $f'(x)$. $$f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 36x + 7) = 6x^2 - 6x - 36.$$

Step 2: Factorize $f'(x)$ and find critical points. $$f'(x) = 6(x^2 - x - 6) = 6(x - 3)(x + 2).$$ Set $f'(x) = 0$: $$6(x - 3)(x + 2) = 0 \implies x = 3 \text{ or } x = -2.$$

Step 3: The points $x = -2$ and $x = 3$ divide the real line into three intervals: $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$.

Step 4: Test the sign of $f'(x)$.

• In $(-\infty, -2)$: Take $x = -3$. $$f'(-3) = 6(-6)(-1) = 36 > 0.$$ So, $f(x)$ is strictly increasing.
• In $(-2, 3)$: Take $x = 0$. $$f'(0) = 6(-3)(2) = -36 < 0.$$ So, $f(x)$ is strictly decreasing.
• In $(3, \infty)$: Take $x = 4$. $$f'(4) = 6(1)(6) = 36 > 0.$$ So, $f(x)$ is strictly increasing.

Answer: The function is strictly increasing in $(-\infty, -2) \cup (3, \infty)$ and strictly decreasing in $(-2, 3)$.

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Example 3: Proving a Function is Strictly Increasing Everywhere

Show that the function $f(x) = x^3 - 3x^2 + 4x, \; x \in \mathbb{R}$ is strictly increasing on $\mathbb{R}$.

Solution: Step 1: Find $f'(x)$. $$f'(x) = 3x^2 - 6x + 4.$$

Step 2: Show that $f'(x) > 0$ for all real values of $x$ by completing the square. $$f'(x) = 3(x^2 - 2x) + 4$$ $$f'(x) = 3[(x-1)^2 - 1] + 4$$ $$f'(x) = 3(x-1)^2 - 3 + 4$$ $$f'(x) = 3(x-1)^2 + 1.$$

Step 3: Analyze the expression. Since $(x-1)^2 \ge 0$ for all real $x$, we get $$3(x-1)^2 \ge 0.$$ Hence, $$3(x-1)^2 + 1 \ge 1 > 0.$$ So, $$f'(x) > 0 \quad \text{for all } x \in \mathbb{R}.$$

Step 4: Conclude using the first derivative test. Since the derivative is positive everywhere, the function is strictly increasing on $\mathbb{R}$.

Answer: Proved analytically.

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Example 4: Trigonometric Function Intervals

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

Solution: Step 1: Find $f'(x)$. $$f'(x) = \cos x - \sin x.$$

Step 2: Find critical points by setting $f'(x) = 0$. $$\cos x - \sin x = 0 \implies \cos x = \sin x \implies \tan x = 1.$$ In the interval $[0, 2\pi]$, this happens at $$x = \frac{\pi}{4}, \; \frac{5\pi}{4}.$$

Step 3: These points divide the interval into three sub-intervals: $$(0, \tfrac{\pi}{4}), \quad (\tfrac{\pi}{4}, \tfrac{5\pi}{4}), \quad (\tfrac{5\pi}{4}, 2\pi).$$

Step 4: Test the sign of $f'(x)$ in each interval.

• In $(0, \frac{\pi}{4})$: Take $x = \frac{\pi}{6}$. $$f'\left(\frac{\pi}{6}\right) = \cos \frac{\pi}{6} - \sin \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{1}{2} > 0.$$ So $f(x)$ is strictly increasing.
• In $(\frac{\pi}{4}, \frac{5\pi}{4})$: Take $x = \frac{\pi}{2}$. $$f'\left(\frac{\pi}{2}\right) = 0 - 1 = -1 < 0.$$ So $f(x)$ is strictly decreasing.
• In $(\frac{5\pi}{4}, 2\pi)$: Take $x = \frac{3\pi}{2}$. $$f'\left(\frac{3\pi}{2}\right) = 0 - (-1) = 1 > 0.$$ So $f(x)$ is strictly increasing.

Answer: Strictly increasing in $(0, \frac{\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi)$ and strictly decreasing in $(\frac{\pi}{4}, \frac{5\pi}{4})$.

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Example 5: Logarithmic Function Intervals

Find the intervals in which the function $f(x) = \ln x - x$ is strictly increasing or strictly decreasing. Also, mention the domain.

Solution: Step 1: Determine the domain. Since $\ln x$ is defined only for $x>0$, the domain is $$(0, \infty).$$

Step 2: Find $f'(x)$. $$f'(x) = \frac{d}{dx}(\ln x - x) = \frac{1}{x} - 1 = \frac{1-x}{x}.$$

Step 3: Find critical points by setting $f'(x)=0$. $$\frac{1-x}{x} = 0 \implies 1-x=0 \implies x=1.$$ This lies in the domain.

Step 4: The domain is divided into two intervals: $(0,1)$ and $(1,\infty)$.

Step 5: Test the sign of $f'(x)$.

• In $(0,1)$: Take $x=\tfrac{1}{2}$. $$f'\left(\tfrac{1}{2}\right)=\frac{1-1/2}{1/2}=1>0.$$ So $f(x)$ is strictly increasing.
• In $(1,\infty)$: Take $x=2$. $$f'(2)=\frac{1-2}{2}=-\frac{1}{2}<0.$$ So $f(x)$ is strictly decreasing.

Answer: Domain: $(0, \infty)$. Strictly increasing in $(0,1)$ and strictly decreasing in $(1,\infty)$.

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Example 6: Exponential Function with Product Rule

Find the intervals in which the function $f(x) = x e^{-x}$ is strictly increasing or strictly decreasing.

Solution: Step 1: Find $f'(x)$ using the product rule. $$f'(x) = x\frac{d}{dx}(e^{-x}) + e^{-x}\frac{d}{dx}(x).$$ Since $\frac{d}{dx}(e^{-x}) = -e^{-x}$, $$f'(x) = x(-e^{-x}) + e^{-x}(1) = e^{-x}(1-x).$$

Step 2: Find critical points by setting $f'(x)=0$. $$e^{-x}(1-x)=0.$$ But $e^{-x}>0$ for every real $x$, so the derivative can be zero only when $$1-x=0 \implies x=1.$$

Step 3: This divides the real line into $(-\infty,1)$ and $(1,\infty)$.

Step 4: Test the sign of $f'(x)$.

• In $(-\infty,1)$: Since $1-x>0$ and $e^{-x}>0$, we get $f'(x)>0$. So the function is strictly increasing.
• In $(1,\infty)$: Since $1-x<0$ and $e^{-x}>0$, we get $f'(x)<0$. So the function is strictly decreasing.

Answer: Strictly increasing in $(-\infty, 1)$ and strictly decreasing in $(1, \infty)$.

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Example 7: Proving Monotonicity Involving Inequalities

Find the least value of $a$ such that the function $f$ given by $f(x) = x^2 + ax + 1$ is strictly increasing on $(1, 2)$.

Solution: Step 1: Find $f'(x)$. $$f'(x) = 2x + a.$$

Step 2: For $f(x)$ to be strictly increasing on $(1,2)$, it is enough to ensure $$f'(x) \ge 0 \text{ for all } x \in (1,2),$$ and in fact for this linear derivative, the least value occurs at the left end of the interval. Since $2x+a$ is a linear expression increasing with $x$, its smallest value on $(1,2)$ is approached as $x \to 1^+$. So we need $$2(1) + a \ge 0 \implies a \ge -2.$$

Step 3: Check the least value $a=-2$. Then $$f'(x)=2x-2=2(x-1).$$ For every $x \in (1,2)$, we have $x-1>0$, so $$f'(x)>0.$$ Thus the function is indeed strictly increasing on $(1,2)$.

Answer: The least value of $a$ is $-2$.