Function Analysis

Class 12 Maths NCERT: Applications of Derivatives

Determining if a Function is Increasing or Decreasing in an Interval

Understanding whether a function is increasing or decreasing over a particular interval is a fundamental application of derivatives. It essentially tells us about the behavior of the function's graph – is it going uphill or downhill?

The core concept here is that the **sign of the first derivative** ($\boldsymbol{f'(x)}$) directly reveals the behavior of the original function ($\boldsymbol{f(x)}$):

• If $f′(x) > 0$ in an interval, the function $f(x)$ is strictly increasing in that interval. This means as $x$ increases, $f(x)$ also increases. 📈
• If $f′(x) < 0$ in an interval, the function $f(x)$ is strictly decreasing in that interval. This means as $x$ increases, $f(x)$ decreases. 📉
• If $f′(x) = 0$ in an interval, the function $f(x)$ is constant in that interval.

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Step-by-Step Solution Methodology

To determine the intervals where a function is increasing or decreasing, follow these systematic steps:

Step 1: Find the First Derivative of the Function ($f′(x)$)

This is the foundational step. You need to differentiate the given function $f(x)$ with respect to $x$. Apply all the differentiation rules you've learned (power rule, product rule, quotient rule, chain rule, etc.).

Example: If $f(x) = x^3 - 3x^2 + 2x$, then:

$$f′(x) = \frac{d}{dx}(x^3 - 3x^2 + 2x) = 3x^2 - 6x + 2$$

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Step 2: Find the Critical Points

Critical points are the values of $x$ where the first derivative $f′(x)$ is either **equal to zero** ($f′(x) = 0$) or **undefined**. These points are crucial because they are the *only* places where the function's increasing/decreasing nature can potentially change.

• Set $f′(x) = 0$: Solve the resulting equation for $x$. These solutions are your primary critical points.
• Check for Undefined $f′(x)$: Look for values of $x$ that would make $f′(x)$ undefined (e.g., division by zero, square root of a negative number). These are also critical points. However, for polynomial functions (which are common in Class 12), $f′(x)$ is usually defined for all real numbers.

Example (continuing from above): For $f′(x) = 3x^2 - 6x + 2$, set $3x^2 - 6x + 2 = 0$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3}$$

So, the critical points are $x = \frac{3 - \sqrt{3}}{3}$ and $x=3+\sqrt{3}3$

x = \frac{3 + \sqrt{3}}{3}.

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Step 3: Determine the Intervals

The critical points divide the function's domain (usually $(-\infty, \infty)$ for most NCERT problems) into several disjoint intervals. These are the intervals you will test.

Example (continuing): The critical points are approximately $x \approx \frac{3 - 1.732}{3} \approx 0.42$ and $x \approx \frac{3 + 1.732}{3} \approx 1.58$.

These divide the number line into three intervals:

• $(-\infty, \frac{3 - \sqrt{3}}{3})$
• $(\frac{3 - \sqrt{3}}{3}, \frac{3 + \sqrt{3}}{3})$
• $(\frac{3 + \sqrt{3}}{3}, \infty)$

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Step 4: Test Each Interval

Choose a **test value** (any number) within each of the intervals determined in Step 3. Substitute this test value into the **first derivative, $f′(x)$**. The sign of $f′(x)$ at this test value will tell you whether the function is increasing or decreasing in that entire interval.

• If $f′(\text{test value}) > 0$, then $f(x)$ is strictly increasing in that interval.
• If $f′(\text{test value}) < 0$, then $f(x)$ is strictly decreasing in that interval.

Example (continuing):

• Interval $(-\infty, \frac{3 - \sqrt{3}}{3})$ (e.g., test $x=0$): $f′(0) = 3(0)^2 - 6(0) + 2 = 2$. Since $2 > 0$, $f(x)$ is strictly increasing in this interval.
• Interval $(\frac{3 - \sqrt{3}}{3}, \frac{3 + \sqrt{3}}{3})$ (e.g., test $x=1$): $f′(1) = 3(1)^2 - 6(1) + 2 = 3 - 6 + 2 = -1$. Since $-1 < 0$, $f(x)$ is strictly decreasing in this interval.
• Interval $(\frac{3 + \sqrt{3}}{3}, \infty)$ (e.g., test $x=2$): $f′(2) = 3(2)^2 - 6(2) + 2 = 12 - 12 + 2 = 2$. Since $2 > 0$, $f(x)$ is strictly increasing in this interval.

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Step 5: Write the Conclusion

Clearly state the intervals where the function is strictly increasing and where it is strictly decreasing based on your findings in Step 4.

Example (continuing):

• $f(x)$ is strictly increasing in $(-\infty, \frac{3 - \sqrt{3}}{3})$ and $(\frac{3 + \sqrt{3}}{3}, \infty)$.
• $f(x)$ is strictly decreasing in $(\frac{3 - \sqrt{3}}{3}, \frac{3 + \sqrt{3}}{3})$.

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Important Notes:

Strictly Increasing/Decreasing vs. Increasing/Decreasing:

• Strictly increasing/decreasing means $f′(x) > 0$ or $f′(x) < 0$. The critical points themselves are usually *excluded* from these intervals, indicated by parentheses $( )$.
• Increasing/decreasing (non-strict) means $f′(x) \ge 0$ or $f′(x) \le 0$. In this case, the critical points (where $f′(x)=0$) *are included* in the intervals, often by using square brackets $[ ]$. NCERT typically asks for "strictly increasing/decreasing."

Polynomial Functions:

For polynomial functions, $f′(x)$ is always defined for all real numbers. So, you only need to set $f′(x) = 0$ to find critical points.

Functions with Restricted Domains:

If the function has a restricted domain (e.g., $\sqrt{x-1}$ has domain $x \ge 1$; $\log x$ has domain $x > 0$), make sure your intervals are entirely within that domain.

Trigonometric Functions:

For trigonometric functions like $f(x) = \sin x$, the critical points (where $f′(x)=\cos x=0$) will often repeat due to periodicity. You usually specify the intervals within a given period (e.g., $(0, 2\pi)$). For $f(x)=\sin x$:

• For $x \in (0, \frac{\pi}{2})$, $\cos x > 0$, so $\sin x$ is strictly increasing.
• For $x \in (\frac{\pi}{2}, \pi)$, $\cos x < 0$, so $\sin x$ is strictly decreasing.
• In $(0, \pi)$, since it increases and then decreases, it is neither strictly increasing nor strictly decreasing over the *entire* interval.