First Derivative Test - Examples, Steps, Applications

First Derivative Test

The First Derivative Test is a fundamental concept in calculus used to determine the monotonicity and local extrema of a function within a specific domain. It helps identify whether a function is increasing, decreasing, or stationary at a particular point by analyzing the sign of its first derivative. This method is widely used in optimization problems and curve sketching in both Class 12 Maths and competitive exams like JEE Main and CUET.

Understanding Monotonicity of a Function

A function is said to be monotonic if it consistently increases or decreases throughout its entire domain. In other words, a monotonic function does not change its direction of increase or decrease.

If a function $f(x)$ is increasing in an interval $I$, then for every pair of points $x_1, x_2 \in I$ such that $x_1 < x_2$, we have $f(x_1) < f(x_2)$.
Similarly, if $f(x)$ is decreasing in $I$, then $f(x_1) > f(x_2)$ for $x_1 < x_2$.

Thus, by a monotonic function $f$ in an interval $I$, we mean that $f$ is either increasing in the given domain or decreasing in the given domain.

Critical Points and Their Role in Extrema

The extrema (maxima and minima) of a function always occur at its critical points. These are points in the domain of a function where either:

• the derivative of the function is zero, or

• the function is not differentiable.

Mathematically, a function $f(x)$ has a critical point at $x = a$ if
$f'(a) = 0$ or $f'(a)$ is not defined.

Critical points are essential for analyzing turning points of a graph, as they indicate where the slope of the tangent line becomes horizontal (i.e., the function stops increasing or decreasing momentarily).

First Derivative Test to Get Extrema

(1) At the Critical Point $x = a$

Let $y = f(x)$ be a differentiable function, and let $x = a$ be a critical point of the function $f(x)$.

1. If $f'(x)$ changes from positive to negative at $x = a$, then $f$ has a local maximum at $x = a$.

2. If $f^{\prime}(x)$ changes from negative to positive at $x=a$, then $f$ has a Local Minima at $\mathrm{x}=\mathrm{a}$.

3. If $f^{\prime}(x)$ does not change any sign at $x=a$, then $f$ has neither Local maxima nor Local minima at $\mathrm{x}=\mathrm{a}$.

(2) At the End Points of the Closed Interval $[a, b]$

Let $f(x)$ be defined on the closed interval $[a, b]$.
If $f'(x) < 0$ for all $x$ in $(a, b)$, then:

• $f(x)$ has a local maximum at $x = a$, and

• $f(x)$ has a local minimum at $x = b$.

Again if $f^{\prime}(x)>0$ then $f(x)$ has a local minimum at $x=a$ and local maximum at $x=b$.

$n^{\text{th}}$ Derivative Test

To determine the nature of stationary points when the first derivative test is inconclusive, we use the $n^{\text{th}}$ derivative test.

Step 1: Find the value of $x$ such that $f'(x) = 0$.
Let at $x = a$, $f'(a) = 0$.

Step 2: Compute higher-order derivatives of $f(x)$ at $x = a$.

1. If $f''(a) < 0$, then $f(x)$ has a local maximum at $x = a$.

2. If $f''(a) > 0$, then $f(x)$ has a local minimum at $x = a$.

3. If $f''(a) = 0$, then proceed to find higher derivatives.

Step 3:
Find $f'''(x)$ at $x = a$.

• If $f'''(a) \neq 0$, then $f(x)$ has neither a maximum nor a minimum at $x = a$ (it is an inflection point).

Step 4:
If $f'''(a) = 0$, then find the fourth derivative $f^{(iv)}(x)$ at $x = a$.

• If $f^{(iv)}(a) < 0$, then $f(x)$ has a local maximum at $x = a$.

• If $f^{(iv)}(a) > 0$, then $f(x)$ has a local minimum at $x = a$.

The process continues similarly for higher-order derivatives if lower ones vanish.

Discontinuous Function

If a function $f(x)$ is discontinuous at $x = a$, but $f(a)$ exists finitely,
we use the basic definition of maxima and minima to determine the nature of the point.

If $f(a) > f(a - h)$ and $f(a) > f(a + h)$, then $x = a$ is a point of local maxima.
If $f(a) < f(a - h)$ and $f(a) < f(a + h)$, then $x = a$ is a point of local minima.

In such cases, the graph of the function can also be used to visually identify the points of maxima and minima.

Non-Differentiable Function

If $f(x)$ is continuous at $x = a$ but not differentiable at $x = a$,
the nature of the point can be checked by observing the change in sign of the derivative near $x = a$.

• If $f'(a - h) > 0$ and $f'(a + h) < 0$, then $x = a$ is a point of local maxima.

• If $f'(a - h) < 0$ and $f'(a + h) > 0$, then $x = a$ is a point of local minima.

Global Maxima and Minima in $[a, b]$

Let $y = f(x)$ be a function defined in a domain $D$, and let $[a, b] \subseteq D$.
The global maxima and global minima of $f(x)$ on the closed interval $[a, b]$ are the greatest and least values of $f(x)$ within this interval.

The global extrema always occur either at the critical points of $f(x)$ in $(a, b)$ or at the endpoints $a$ and $b$.

To find them:

1. Determine all critical points $c_1, c_2, \ldots, c_n$ in $(a, b)$ where $f'(x) = 0$, or where $f(x)$ is non-differentiable or discontinuous.

2. Evaluate $f(x)$ at these points and at the endpoints of the interval.

Let the corresponding function values be:
$f(c_1), f(c_2), \ldots, f(c_n)$, along with $f(a)$ and $f(b)$.

Now define:

$M_1 = \max {f(a), f(c_1), f(c_2), \ldots, f(c_n), f(b)}$
$M_2 = \min {f(a), f(c_1), f(c_2), \ldots, f(c_n), f(b)}$

Then,

• $M_1$ represents the Global Maximum (Absolute Maximum) or Greatest Value of the function.

• $M_2$ represents the Global Minimum (Absolute Minimum) or Least Value of the function.

Global Maxima and Minima in $(a, b)$

Let us consider a function $f$ given by $f(x) = x + 2$, where $x \in (0, 1)$.
Observe that this function is continuous on $(0, 1)$, yet it neither has a maximum nor a minimum value within this open interval. In fact, the function does not even have local maxima or minima in $(0, 1)$.

However, if we extend the domain of $f$ to the closed interval $[0, 1]$, the situation changes.
Although the function still lacks local extrema, it now has definite maximum and minimum values within the interval.

Here,
$f(1) = 3$ is the maximum value of $f$ at $x = 1$, and
$f(0) = 2$ is the minimum value of $f$ at $x = 0$.

Thus,

• The maximum value $3$ at $x = 1$ is called the absolute maximum value (also known as the global maximum or greatest value) of $f$ on $[0, 1]$.

• The minimum value $2$ at $x = 0$ is called the absolute minimum value (also known as the global minimum or least value) of $f$ on $[0, 1]$.

Understanding Through Graph

Consider the graph of a continuous function defined on a closed interval $[a, d]$.

From the graph:

• The function $f$ has a local minimum at $x = b$, with the local minimum value $f(b)$.

• It also has a local maximum at $x = c$, with the local maximum value $f(c)$.

• Additionally, the function attains an absolute maximum at $x = a$, with value $f(a)$, and an absolute minimum at $x = d$, with value $f(d)$.

It is important to note that the absolute (global) extrema of a function may differ from its local extrema.
Local maxima and minima depend on the neighborhood of the point, while global maxima and minima represent the overall highest and lowest values of the function across its entire domain.

Method for Finding the Greatest and Least Values of $f(x)$ in $(a, b)$

The method for finding the greatest and least values of a function $f(x)$ in an open interval $(a, b)$ is almost the same as the process used for a closed interval $[a, b]$, but requires a little extra care.

Step 1

We do not take $f(a)$ and $f(b)$ into consideration in the first step.

Let $M_1 = \max { f(c_1), f(c_2), \ldots, f(c_n) }$ and $M_2 = \min { f(c_1), f(c_2), \ldots, f(c_n) }$ where $c_1, c_2, \ldots, c_n$ are the critical points of $f(x)$ in $(a, b)$.

Step 2

If $\lim_{x \to a^{+}} f(x) > M_1$ or $\lim_{x \to b^{-}} f(x) > M_1$, then $f(x)$ does not possess a global maximum.

If none of these two conditions are true, then $M_1$ is the global maximum.

Similarly, if $\lim_{x \to a^{+}} f(x) < M_2$ or $\lim_{x \to b^{-}} f(x) < M_2$, then $f(x)$ does not possess a global minimum.

If neither of these conditions are satisfied, then $M_2$ is the global minimum.

For example,

This function has no global maxima and no global minima.

Comparison of Derivative Tests

This section provides a side-by-side comparison to understand how each test determines the nature of critical points based on the behaviour of derivatives. This section will also help you see when one test is more effective or simpler to apply than the others.

Application of the First Derivative Test

The First Derivative Test uses these critical points to determine:

• Whether the function is increasing or decreasing in a given interval.

• Whether a point corresponds to a local maximum or local minimum.

By analyzing the sign change of $f'(x)$ around a critical point:

• If $f'(x)$ changes from positive to negative, the function has a local maximum.

• If $f'(x)$ changes from negative to positive, the function has a local minimum.

• If there is no sign change, the point is neither a maximum nor a minimum.

This test forms the backbone of curve analysis and optimization problems in calculus, making it a crucial topic for students preparing for board exams, JEE, and other entrance tests.

Solved Examples Based on First Derivative Test

Example 1: If the function $f$ given by $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$, for some $a \in R$, is increasing in $(0,1]$ and decreasing in $[1,5)$, then one root of the equation $\frac{f(x) - 14}{(x - 1)^2} = 0, (x \neq 1)$ is:

1. 7

2. 6

3. 5

4. −7

Solution:

$f'(x) = 3(x^2 - 2(a - 2)x + a)$
$f'(1) = 0$
$\Rightarrow a = 5$
$\frac{f(x) - 14}{(x - 1)^2} = \frac{x^3 - a x^2 + 15x - 7}{(x - 1)^2} = x - 7 = 0$
$\Rightarrow x = 7$

Hence, the answer is option 1.

Example 2: The maximum value of the term independent of $t$ in the expansion of $\left(tx^{\frac{1}{5}} + \frac{(1 - x)^{\frac{1}{10}}}{t}\right)^{10}$ where $x \in (0,1)$ is:

1. $\frac{2 \cdot 10!}{3\sqrt{3}(5!)^2}$

2. $\frac{10!}{3(5!)^2}$

3. $\frac{10!}{\sqrt{3}(5!)^2}$

4. $\frac{2 \cdot 10!}{3(5!)^2}$

Solution:
The term independent of $t$ will be the middle term, as it has equal magnitude but opposite powers of $t$.

So,
$T_6 = {}^{10}C_5 (tx^{\frac{1}{5}})^5 \left(\frac{(1 - x)^{\frac{1}{10}}}{t}\right)^5$
$T_6 = f(x) = {}^{10}C_5 (x \sqrt{1 - x})$
$f'(x) = {}^{10}C_5 \left(\sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}}\right) = {}^{10}C_5 \left(-\frac{3x - 2}{2\sqrt{1 - x}}\right)$

For maximum, $f'(x) = 0$
$\Rightarrow x = \frac{2}{3}$
$f''(x) = \frac{2 - 3x}{4(1 - x)^{\frac{3}{2}}} - \frac{3}{2\sqrt{1 - x}} = \frac{3x - 4}{4(1 - x)^{\frac{3}{2}}}$
$f''\left(\frac{2}{3}\right) < 0$

So, $f(x)_{\max} = {}^{10}C_5 \left(\frac{2}{3}\right) \cdot \frac{1}{\sqrt{3}}$

Hence, the answer is option 1.

Example 3: The function $f$ defined by $f(x) = x^3 - 3x^2 + 5x + 7$ is: [JEE Main 2017]

1. increasing in $R$

2. decreasing in $R$

3. decreasing in $(0, ∞)$ and increasing in $(−∞, 0)$

4. increasing in $(0, ∞)$ and decreasing in $(−∞, 0)$

Solution:

First Derivative Test to get extrema:

Find $x$ such that $f'(x) = 0$. Let $f'(a) = 0$ at $x = a$.

If $f'(a) < 0$, then $f(x)$ is maximum at $x = a$.
If $f'(a) > 0$, then $f(x)$ is minimum at $x = a$.
If $f'(a) = 0$, check higher derivatives.

$f(x) = x^3 - 3x^2 + 5x + 7$
$f'(x) = 3x^2 - 6x + 5$
$D = (6)^2 - 4(3)(5) = 36 - 60 < 0$

So, $f'(x) > 0$ for all $x \in R$

Hence, the answer is option 1.

Example 4: Let $f(x) = \frac{x}{\sqrt{a^2 + x^2}} - \frac{d - x}{\sqrt{b^2 + (d - x)^2}}, x \in R$ where $a, b,$ and $d$ are non-zero real constants. [JEE Main 2019]

1. $f$ is neither increasing nor decreasing function of $x$

2. $f'$ is not a continuous function of $x$

3. $f$ is a decreasing function of $x$

4. $f$ is an increasing function of $x$

Solution:

$f'(x) = \frac{a^2}{(a^2 + x^2)^{\frac{3}{2}}} + \frac{b^2}{(b^2 + (d - x)^2)^{\frac{3}{2}}} > 0, , x \in R$

Hence, $f(x)$ is an increasing function, so the answer is option 4.

Example 5: Let $f:[0,2] \rightarrow \mathbb{R}$ be a twice differentiable function such that $f''(x) > 0$ for all $x \in (0,2)$. If $\phi(x) = f(x) + f(2 - x)$, then $\phi$ is: [JEE Main 2019]

1. decreasing on $(0,1)$ and increasing on $(1,2)$

2. increasing on $(0,2)$

3. increasing on $(0,1)$ and decreasing on $(1,2)$

4. decreasing on $(0,2)$

Solution:

Given $f''(x) > 0$, which means $f'(x)$ is increasing.
Now, $\phi'(x) = f'(x) - f'(2 - x)$

For $x \in (0,1)$, $f'(x) - f'(2 - x) < 0 \Rightarrow \phi'(x) < 0$
For $x \in (1,2)$, $f'(x) - f'(2 - x) > 0 \Rightarrow \phi'(x) > 0$

Hence, $\phi(x)$ is decreasing on $(0,1)$ and increasing on $(1,2)$.
Answer: option 1.

List of topics related to First Derivative Test

This section provides an overview of all the key calculus topics linked to the First Derivative Test, such as continuity, differentiability, and behavior of functions around critical points. It helps learners understand how the sign of the first derivative determines local maxima and minima.

Differentiability and Existence of Derivative

Examining differentiability Using Graph of Function

Continuity of Composite Function

Continuity And Differentiability

NCERT Resources

This section compiles authentic NCERT Class 12 Maths study materials that align with Chapter 5, Continuity and Differentiability. It includes notes, solved exercises, and exemplar problems that build conceptual clarity for applying the First Derivative Test effectively.

NCERT Class 12 Maths Notes for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Solutions for Chapter 5 - Continuity and Differentiability

NCERT Class 12 Maths Exemplar Solutions for Chapter 5 - Continuity and Differentiability

Practice Questions based on First Derivative Test

Sharpen your skills with conceptual and numerical MCQs focused on identifying maxima, minima, and critical points using the First Derivative Test. These practice questions are perfect for board exams and entrance preparation.

First Derivative Test- Practice Question MCQ

We have shared below the links to practice questions on the related topics to First Derivative Test: