Inflection Point: Definition, Graph and Example

Understanding Concavity and Convexity of a Function

Concavity describes how a curve bends — whether it opens upward or downward. It is determined by the second derivative of a function. Knowing concavity helps in identifying the shape of the graph and locating points of inflection or curvature change.

Concave Up (Convex Down)

If $f''(x) > 0$ in an interval $(a, b)$, the curve of $f(x)$ is said to be concave upward (or convex down).
In this case, the slope of the tangent increases as $x$ increases, and the function appears to bend upward when viewed from the x-axis.

Mathematically:
$
f''(x) > 0 \text{ for } x \in (a, b) \Rightarrow f(x) \text{ is concave upward.}
$

Concave Down (Convex Up)

If $f''(x) < 0$ in an interval $(a, b)$, the curve of $f(x)$ is said to be concave downward (or convex up).
Here, the slope of the tangent decreases as $x$ increases, and the graph bends downward as seen from the x-axis.

Mathematically:
$
f''(x) < 0 \text{ for } x \in (a, b) \Rightarrow f(x) \text{ is concave downward.}
$

Point of Inflection – Definition and Concept

A point of inflection is a point on the graph of a function where the curve changes its concavity — that is, from concave up to concave down or vice versa.
At this point, the second derivative $f''(x)$ is equal to zero, but importantly, it must change sign on either side of that point.

Mathematically:
$
f''(a) = 0 \text{ and } f''(x) \text{ changes sign at } x = a.
$

If the function transitions from concave to convex (or vice versa) at $x = a$, then $x = a$ is a point of inflection.

Inflection Point and Derivative Behavior

If $f'(x)$ (the first derivative) does not change sign as $x$ increases through $c$, then $x = c$ is neither a point of local maxima nor minima.
Instead, such a point is referred to as a point of inflection, indicating that the function is still increasing or decreasing but with a change in curvature.

In summary:

• When $f''(x) = 0$ and $f''(x)$ changes sign → Point of Inflection.

• When $f'(x)$ does not change sign → No extremum (the function continues increasing or decreasing).

Behavior of Tangent and Second Derivative at the Point of Inflection

At the point of inflection, the tangent drawn to the smooth curve at $x = a$ actually crosses the curve — unlike at maxima or minima, where the tangent merely touches the curve.

The second derivative, $f''(x)$, changes its sign in the neighborhood of the point of inflection. Therefore, a necessary condition for a point of inflection is that:

$
f''(x) = 0 \text{ (if it exists) at } x = a
$

However, the mere fact that $f''(x) = 0$ at some point does not guarantee that the point is an inflection point. A sign change in $f''(x)$ around that point is essential.

Sufficient Condition for a Point of Inflection

For a point $x = a$ to qualify as a true point of inflection, a sufficient condition is required.
The function $f''(a - h)$ and $f''(a + h)$ must have opposite signs in a small neighborhood around $x = a$.

Mathematically,
$
f''(a - h) \cdot f''(a + h) < 0
$

This condition ensures that the curve actually changes concavity (from concave up to concave down or vice versa) near $x = a$.

Example – Inflection Point in $f(x) = x^3$

Let’s consider the function $f(x) = x^3$.

We have:
$
f'(0) = 0, \quad f''(0) = 0
$

Also,
$
f''(x) < 0 \text{ for } x < 0, \quad f''(x) > 0 \text{ for } x > 0
$

Thus, $f''(x)$ changes sign as $x$ crosses $0$, meaning $f(x)$ changes its concavity at $x = 0$.

Therefore, $x = 0$ is a point of inflection, where the curve transitions from being concave downward to concave upward.

General Condition for Inflection

A point $x = a$ is a point of inflection only if:

1. $f''(x)$ changes sign as $x$ passes through $a$.

2. Only those values of $x$ for which $f''(x)$ actually changes sign qualify as true inflection points.

For example:
$
f''(x) = x^2(x - 2)
$

Here,
$
f''(x) = 0 \text{ at } x = 0 \text{ and } x = 2
$

But only $x = 2$ is a point of inflection, because at $x = 2$, $f''(x)$ actually changes sign.

Example – No Inflection at $x = 0$ for $f(x) = x^4$

Consider:
$
f(x) = x^4, \quad f''(x) = 12x^2
$

We have:
$
f''(0) = 0
$

However, since $f''(x) = 12x^2$ does not change sign in $(0 - h, 0 + h)$, the point $x = 0$ is not an inflection point.
This shows that $f''(x) = 0$ is only a necessary, not a sufficient, condition.

Example – Inflection Without Differentiability

Let:
$
f(x) = 2x^{1/3}
$

Then,
$
f'(x) = \frac{2}{3}x^{-2/3}
$

Here, $f'(x)$ is not differentiable at $x = 0$ because the graph has a vertical tangent at that point.
However, the graph of $f(x)$ changes concavity at $x = 0$.

Hence, even though $f(x)$ is not differentiable at $x = 0$, the point $x = 0$ is still a point of inflection.
This proves that differentiability is not always necessary for the existence of an inflection point.

NOTE:

It is not necessary that at the point of inflection, the function is continuous.

Solved Examples Based on Concavity and Point of Inflection

Example 1: $f(x)=x^3-3 x^2$ has concaity upwards in the interval
1) $(1, \infty)$
2) $(-\infty, \infty)$
3) $(-1, \infty)$
4) $(-\infty, 1)$

Solution:
As we have learned
Concavity, Convexity, of a function -
For concavity:
If $f^{\prime \prime}(x)>0$ in the interval $(a, b)$ then shape of $\mathrm{f}(\mathrm{x})$ in interval $(a, b)$ is concave when observed from upwards or convex down.

For convexity:
If $f^{\prime \prime}(x)<0$ in the interval $(a, b)$ then it is convex upward or concave down.
- wherein

$
f^{\prime}(x)=3 x^2-6 x \Rightarrow f^{\prime \prime}(x)=6 x-6=6(x-1)
$

for concave up $\rightarrow f^{\prime \prime}(x)>0 \Rightarrow 6(x-1)>0 \Rightarrow x>1$

Example 2: $f(x)=x^4+2 x^3+6 x^2+12 x$ has concavity upwards only in the interval
1) $(1, \infty)$
2) $(-\infty, 0)$
3) $(0, \infty)$
4) $(-\infty, \infty)$

Solution:
As we have learned
Concavity, Convexity, of a function -
For concavity:
If $f^{\prime \prime}(x)>0$ in the interval $(a, b)$ then shape of $\mathrm{f}(\mathrm{x})$ in interval $(a, b)$ is concave when observed from upwards or convex down.

For convexity:
If $f^{\prime \prime}(x)<0$ in the interval $(a, b)$ then it is convex upward or concave down.
- wherein

$
\begin{aligned}
& f^{\prime}(x)=4 x^3+6 x^2+12 x+12 \\
& f^{\prime \prime}(x)=12 x^2+12 x+12=12\left(x^2+x+1\right)>0 \forall n \epsilon R \\
& \because f^{\prime \prime}(x)>0 \ldots . \forall x \epsilon R \\
& \Rightarrow f(x) \text { is concave up } \forall n \in R
\end{aligned}
$

Example 3: Which of the following curves shows $x=a$ as point of inflection?

1)

2)

3)

4)

Solution:

As we have learned

Point of inflection -

If at $x=a$ the shape of the curve changes from concave to convex from convex to concave then at $\mathrm{x}=\mathrm{a}$ is known as the point of inflection and
$
f^{\prime \prime}(x)=0 \quad \text { at } x=a
$
In (A), concavity remains downward in left and right both of '$a$'
In (B), concavity changes from downward to upward about '$a$'
In (C), concavity remains downward about '$a$'

Example 4: Which of the following is not a point of inflection in $y=\sin x$
1) $\frac{\pi}{2}$
2) $\pi$
3) $2 \pi$
4) $3 \pi$

Solution:
As we have learned
Point of inflection -
If at $x=a$ the shape of the curve changes from concave to convex from convex to concave then at $\mathrm{x}=\mathrm{a}$ is known as the point of inflection and

From the graph we see, concavity remains downward about $x=\pi / 2$ but about $\pi, 2 \pi, 3 \pi$ it changes, so $x=\pi / 2$ is not a point of inflection

Example 5: Point of inflection for $f(x)=x^3$ is at $x$ equals
1) $ -1$
2) $\frac{1}{2}$
3)$0$
4) $1$

Solution:
As we have learned
Condition for point of inflection-
1. $f^{\prime \prime}(x)$ changes sign $x$ passes through the point a.
2. only those values of $x$ for $f^{\prime \prime}(x)$ change signs are points of inflection.
e.g. : $f^{\prime \prime}(x)=x^2(x-2), \quad f^{\prime \prime}(x)=0$
at $x=0$ and $x=2$ but only $x=2$
point of inflection because at $x=2 f^{\prime \prime}(x)$ changes sign only
$f^{\prime}(x)=3 x^2 \Rightarrow f^{\prime \prime}(x)=6 x$ which will change its sign at $\mathrm{x}=0$, so concavity will change at $x=0$, so $x=0$ is a point of inflection.

Hence, the answer is the option 3.

List of topics related to Inflection Point

This section gives an overview of all major subtopics connected to the concept of the inflection point, including differentiability, existence of derivative, continuity of composite function, etc.

Differentiability and Existence of Derivative

Examining differentiability Using Graph of Function

Continuity of Composite Function

Continuity And Differentiability

NCERT Resources

This section compiles important NCERT Class 12 Maths resources related to Chapter 5, helping students reinforce their conceptual clarity through authentic material.

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 Inflection Point

This section offers MCQs and practice questions designed to test your understanding of inflection points, concavity, and curvature concepts.

Inflection Point- Practice Question MCQ

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