Determine The Intervals On Which The Function Is Strictly Increasing

Understanding how to determine the intervals on which the function is strictly increasing is a fundamental skill in calculus and mathematical analysis. This concept allows us to analyze the behavior of functions, optimize real-world scenarios, and sketch accurate graphs. Whether you are a student preparing for an exam or a professional applying mathematical models, mastering this process provides a powerful lens for interpreting change.

The Core Concept: What Does "Strictly Increasing" Mean?

Before diving into the mechanics, Make sure you define the terminology precisely. Now, it matters. A function $f(x)$ is considered strictly increasing on an interval $I$ if, for any two numbers $x_1$ and $x_2$ in $I$ where $x_1 < x_2$, the inequality $f(x_1) < f(x_2)$ holds true.

Visually, this means the graph moves upward as you trace it from left to right. There are no flat sections (horizontal tangents) and no downward dips within that specific interval. This distinction is critical: a function that is merely "increasing" (non-decreasing) allows for flat segments where the derivative equals zero, whereas a strictly increasing function requires the output values to be distinct for distinct inputs.

The Primary Tool: The First Derivative Test

The most efficient method to determine the intervals on which the function is strictly increasing involves the first derivative, denoted as $f'(x)$ or $\frac{dy}{dx}$. The derivative represents the instantaneous rate of change or the slope of the tangent line at any given point.

The logic follows directly from the definition of the derivative:

• If $f'(x) > 0$ for all $x$ in an interval, the slope is positive everywhere. Think about it: * If $f'(x) < 0$, the function is strictly decreasing. Here's the thing — consequently, the function is strictly increasing on that interval. * If $f'(x) = 0$, the function has a horizontal tangent (a critical point), which acts as a potential boundary between increasing and decreasing intervals.

Important Nuance: While $f'(x) > 0$ guarantees a strictly increasing function, the converse is not always strictly true in advanced analysis (e.g., $f(x) = x^3$ is strictly increasing on $\mathbb{R}$, but $f'(0) = 0$). On the flip side, for standard calculus curriculum problems involving polynomials, rational functions, and trigonometric functions, testing where $f'(x) > 0$ is the standard, accepted procedure The details matter here. That alone is useful..

Step-by-Step Procedure

To systematically find these intervals, follow this structured workflow:

1. Find the Derivative

Compute $f'(x)$ using differentiation rules (Power Rule, Product Rule, Quotient Rule, Chain Rule, etc.). Ensure the derivative is simplified as much as possible—factored form is usually best for the next steps.

2. Identify Critical Numbers

Find all values of $x$ where $f'(x) = 0$ or where $f'(x)$ does not exist (undefined).

• Zeros of the derivative: Solve $f'(x) = 0$.
• Points of non-differentiability: Look for sharp corners, cusps, vertical tangents, or domain restrictions (like division by zero in rational functions).

These critical numbers partition the domain of the function into distinct test intervals Not complicated — just consistent..

3. Create a Sign Chart (or Test Points)

Draw a number line and mark the critical numbers. Select a single test point from each interval created by these markers. Plug each test point into the factored derivative $f'(x)$. You do not need the exact numerical value; you only need the sign (positive or negative).

4. Determine the Intervals

• If $f'(test\ point) > 0$, the function is strictly increasing on that entire interval.
• If $f'(test point) < 0$, the function is strictly decreasing.

5. Write the Answer in Interval Notation

Express the final answer using standard interval notation (parentheses for open intervals, brackets for closed intervals if the endpoint is included in the domain and the function is continuous there). Typically, for "strictly increasing," we use open intervals $(a, b)$ because the derivative is zero at the critical points $a$ and $b$, meaning the function is not strictly increasing at those exact points.

Worked Example: Polynomial Function

Let’s apply the steps to a concrete example: $f(x) = x^3 - 3x^2 - 9x + 5$.

Step 1: Find the Derivative $f'(x) = 3x^2 - 6x - 9$

Step 2: Find Critical Numbers Set $f'(x) = 0$: $3x^2 - 6x - 9 = 0$ Divide by 3: $x^2 - 2x - 3 = 0$ Factor: $(x - 3)(x + 1) = 0$ Critical numbers are $x = -1$ and $x = 3$. The derivative exists for all real numbers, so no extra critical points from non-differentiability Simple, but easy to overlook..

Step 3: Sign Chart Intervals to test: $(-\infty, -1)$, $(-1, 3)$, $(3, \infty)$.

You'll probably want to bookmark this section Not complicated — just consistent..

Step 4: Conclusion The function is strictly increasing on the intervals $(-\infty, -1)$ and $(3, \infty)$ Less friction, more output..

Worked Example: Rational Function

Rational functions introduce domain restrictions (vertical asymptotes) which act as additional boundaries for intervals. Consider $f(x) = \frac{x}{x^2 + 1}$ That's the whole idea..

Step 1: Derivative (Quotient Rule) $f'(x) = \frac{(1)(x^2+1) - x(2x)}{(x^2+1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}$

Step 2: Critical Numbers

• Numerator $= 0 \Rightarrow 1 - x^2 = 0 \Rightarrow x = \pm 1$.
• Denominator $= 0 \Rightarrow (x^2+1)^2 = 0$. No real solutions. Domain is all real numbers.

Critical numbers: $x = -1, 1$.

Step 3: Sign Chart The denominator $(x^2+1)^2$ is always positive. The sign of $f'(x)$ depends entirely on the numerator $1 - x^2 = (1-x)(1+x)$ Small thing, real impact..