In Calculus, the graphs of the functions have curves. The direction of the curve defines the words concavity and inflection points. If the graph of a curve is bent upward, like an upright bowl or U shape, then the curve is said to be concave up or convex. If the graph of a curve is bent down, like an inverted U, then the curve is said to be concave down or simply concave.

For any function f(x), If f(x) and f'(x) are both differentiable functions, then f(x) is said to be concave up if f"(x) $\geq$ 0 and concave down if f"(x) $\leq$ 0. The point at which the graph changes its direction of concavity is called Inflection point and at this point f”(x) = 0.

According to the inflection point theorem,
"If f'(a)exists and f "(a) changes sign at x = a, then (a, f(a)) is an inflection point and if f"(a) exists then f"(a) = 0.
This theorem states a necessary but not sufficient condition for x to be an inflection point for f. From the definition, it can also be stated that the sign of f'(x) must be the same on either side of the point (x, y). If the sign is positive, then the point is a rising point of inflection and if the sign is negative, then the point is a falling point of inflection.
Inflection points can also be categorized based on whether f'(x) is zero or not zero.

• If f'(x) = 0, then the point is called a stationary point of inflection, and is also called as a saddle-point.
• If f'(x) $\neq$ 0, then the point is a non-stationary point of inflection.

Look at the graph above. It represents the graph of y = $x^3$. At (0, 0) the tangent becomes the x-axis and cuts the graph at this point. This point (0, 0) is called the stationary point of inflection or saddle point.

Rotate the graph y = $x^3$ slightly about the origin (say about 20° in anticlockwise direction). At (0, 0) the tangent still cuts the graph in two, but its slope is non-zero. This is non-stationary point of inflection.
In short, inflection point can be defined as:
An inflection point on f(x) occurs at ‘a’ if and only if f(x) has a tangent line at ‘a’ and there exists an interval I containing ‘a’ such that f(x) is concave up on one side of ‘a’ and concave down on the other side.

## Inflection Point Graph

In differential calculus, an inflection is a point on a curve at which the curvature or concavity changes sign from plus to minus or from minus to plus. The curve changes from being concave up to concave down, or vice versa.

Consider the graph of the general quadratic function given by f(x)= $ax^2 + bx + c$. We can analyze the concavity by finding the first and second derivatives of the function f(x), we get f'(x) = 2 a x + b and f"(x)=2a.
The sign of ‘a’ indicates the sign of f "(x). The graph of f represents a parabola and it will be concave up if ‘a’ is positive and concave down if ‘a’ is negative.

In the above graph we have f(x) = $x^2 + x + 1$ which is a quadratic function with ‘a’ as positive and g(x) = -$x^2$+x+1 with ‘a’ as negative. We can see that the graphs of f(x) is concave up and that of g(x) is concave down.

Consider one more example and let us describe the concavity and inflection point of the graph of
f(x) = 2$sin^2x$ - $x^2$ for x $\epsilon$ [0, $\pi$]. The functions f(x) and f’(x) are both differentiable.

Thus,
f'(x) = 4 sinx cosx - 2x = 2 sin (2x) – 2x,
And
f"(x) = 4cos(2x)-2.

f"(x) $\geq$ 0 when 0 $\leq x \leq$ $\frac{\pi}{6}$.

Therefore the graph is concave up on [0, $\frac{\pi}{6}$]

f"(x) $\leq$ 0 when $\frac{\pi}{6}$ $\leq$ x $\leq$ $\pi$.

Therefore the graph is concave down on [$\frac{\pi}{6}$, $\pi$]

By solving f”(x) = 0, which is 4 cos(2x) – 2 = 0,

we get only one inflection point as $\frac{\pi}{6}$ in [0, $\pi$].

## Finding Inflection Points

### Solved Examples

Question 1: Find the point of inflection on the curve of f(x) = $4x^3 - 12x^2 + 3x – 7$
Solution:

The derivative of the given function f(x) is f'(x) = $12x^2 - 24x + 3$

The second derivative of the function f(x), f"(x) = 24x - 24

Solving f”(x) = 0, 24x-12 = 0

We get, x = $\frac{24}{24}$ = 1.

f(1) = $4(1)^3 - 12(1)^2 + 3(1) - 7$ = -12,

So (1, -12) is a point of inflection on the curve.

Question 2: Find α and β so that the function f(x) = $αx^3 +β x^2 +1$ has a point of inflection at (-2, 3).
Solution:

The function f(x) is everywhere continuous and differentiable. So we have,
f'(x) = $3αx^2 + 2βx$, and f"(x) = 6αx + 2β.

Given that the point (-2, 3) is on the graph, so f(-2) = $3α(-2)^2 + 2β(-2)$ = 3 $\rightarrow$ 12α - 4β = 3

Also given that the point (-2, 3) is an inflection point, so f ''(-2) = 0 $\rightarrow$ 6α(-2) + 2β = 0$\rightarrow$-12α + 2β = 0

Solving the two equations 12α - 4β = 3 and -12α + 2β = 0, we get α = $\frac{-1}{4}$ and β = -$\frac{3}{2}$.

Thus, we have (f(x) = -($\frac{1}{4}$) $x^3$ +(-$\frac{3}{2}$) $x^2$ + 1) which has inflection point as (-2, 3).

Question 3: Find all inflection points of function f defined by f(x) = $2 x^4 – x^3 + 4$
Solution:

To determine the points of inflection of the given function f(x), we need to find the second derivative f " (x) and examine its sign. This shows the concavity of the graph of f(x) and therefore all points of inflection if any.

f(x) and f'(x) is differentiable, thus we have
f'(x) = $8 x^3 - 3 x^2$

f"(x) = $24 x^2 - 6 x$ = 6x(4x - 1)

The sign of f”(x) depends on the signs of 6x and 4x-1 as f’(x) is the product of 6x and 4x-1. The table below helps us to understand the signs of 6x and 4x - 1 and that of f"(x). Also the concavity can be described from it.

-∞ to 00 to $\frac{1}{4}$$\frac{1}{4}$ to ∞
6x
-
+
+
4x-1
-
-
+
f "(x)
+
-
+
f(x)
Concave upConcave downConcave up

The points of inflection are located where there is a change in concavity of a function.

Therefore the points, (0 ,f(0))= (0 ,4) and ($\frac{1}{4}$, f($\frac{1}{4}$)) = ($\frac{1}{4}$, $\frac{1022}{256}$) = (0.25,3.99) are points of inflection of f(x) = $2 x^4 – x^3 + 4$.