# Inflection Point

How to find the inflection point of a function: definitions, 1 example, and its solution.

## Second Derivative and Graph

### f''(x) is Plus

If f''(x) is plus,

then f'(x), the slope of f(x), increases.

Second Derivative

So if f'(x) is plus and f''(x) is plus,

then f'(x) increases like 1, 2, 3, 4, ... .

So the graph of y = f(x) shows this shape: _{↗}^{↑}.

It's called [concave up].

If f'(x) is minus and f''(x) is plus,

then f'(x), the slope of f(x), increases like -4, -3, -2, -1, ... .

So the graph of y = f(x) shows this shape: ^{↘}_{→}.

It's also called [concave up].

### f''(x) is Minus

If f''(x) is minus,

then f'(x), the slope of f(x), decreases.

So if f'(x) is plus and f''(x) is minus,

then f'(x) decreases like 4, 3, 2, 1, ... .

So the graph of y = f(x) shows this shape: _{↗}^{→}.

It's called [concave down].

If f'(x) is minus and f''(x) is minus,

then f'(x) decreases like -1, -2, -3, -4, ... .

So the graph of y = f(x) shows this shape: ^{↘}_{↓}.

It's also called [concave down].

## Inflection Point

### Definition

The inflection point is the point

where the graph of y = f(x) changes

either from concave up to concave down

or from concave down to concave up.

So it's the point where

f''(x) changes

either from plus to minus

or from minus to plus.

In most cases,

f''(x) is continuous.

So the inflection point is where

f''(x) = 0.

(brown point)

## Example

### Example

### Solution

To find the local maximum and minimum points,

first find the zeros of y = f'(x).

f(x) = x^{3} - 3x^{2} + 6

Then f'(x) = 3x^{2} - 3⋅2x^{1}.

Derivative of a Polynomial

Factor f'(x) = 3x^{2} - 3⋅2x^{1}.

Then f'(x) = 3x(x - 2).

Find the zeros of 3x(x - 2) = 0.

Then x = 0, 2.

f'(x) = 3x(x - 2)

So y = f'(x) is y = 3x(x - 2).

And the zeros are 0 and 2.

Draw y = f'(x).

y = f'(x) is a parabola

that is opened upward

and that passes through 0 and 2.

Next, to find the zeros of y = f''(x),

find f''(x).

f'(x) = 3x^{2} - 3⋅2x^{1}

Then f''(x) = 3⋅2x^{1} - 3⋅2.

f''(x) = 6(x - 1)

So y = f'(x) is y = 6(x - 1).

The zero is 1.

f'(x) = 6(x - 1)

So y = f'(x) is y = 6(x - 1).

And the zero is 1.

Draw y = f'(x).

y = f'(x) is line

that passes through (1, 0).

Point-Slope Form

Make a table like this.

See the graph of y = f'(x)

and fill the f'(x) row.

For x = 0 and 2,

f'(x) = 0.

For x < 0,

f'(x) is plus.

For 0 < x < 2,

f'(x) is minus.

For x > 2,

f'(x) is plus.

See the graph of y = f''(x)

and fill the f''(x) row.

For x = 1,

f''(x) = 0.

For x < 1,

f''(x) is minus.

For x > 1,

f''(x) is plus.

Fill the f(x) row.

If f'(x) is plus and f''(x) is minus,

f(x) shows _{↗}^{→}.

If f'(x) is minus and f''(x) is minus,

f(x) shows ^{↘}_{↓}.

If f'(x) is minus and f''(x) is plus,

f(x) shows ^{↘}_{→}.

If f'(x) is plus and f''(x) is plus,

f(x) shows _{↗}^{↑}.

Then see the table.

At x = 0,

f'(x) changes from upward (_{↗}^{→}) to downward (^{↘}_{↓}).

So x = 0 is the local maximum.

At x = 1,

f''(x) changes from minus to plus.

So x = 1 is the inflection point.

At x = 2,

f(x) changes from downward (^{↘}_{→}) to upward (_{↗}^{↑}).

So x = 2 is the local minimum.

Then find the y values of these points.

Put x = 0, 1, and 2 into the given f(x).

f(0) = 6

f(1) = 4

f(2) = 2

Write these y values into the table.

At (0, 6),

f(x) changes from upward (_{↗}^{→}) to downward (^{↘}_{↓}).

So (0, 6) is the local maximum.

At (1, 4),

f''(x) changes from minus to plus.

So (1, 4) is the inflection point.

At (2, 2),

f(x) changes from downward (^{↘}_{→}) to upward (_{↗}^{↑}).

So (2, 2) is the local minimum.

So (0, 6) is the local maximum.

(2, 2) is the local minimum.

And (1, 4) is the inflection point.

### Graph

To graph y = f(x),

use the table above.

For x < 0,

f(x) shows _{↗}^{→}.

(0, 6) is the local maximum.

For 0 < x < 1,

f(x) shows ^{↘}_{↓}.

(1, 4) is the inflection point.

The graph changes

from concave down (^{↘}_{↓}) to concave up (^{↘}_{→}).

For 1 < x < 2,

f(x) shows ^{↘}_{→}.

(2, 2) is the local minimum.

For x > 3,

f(x) shows _{↗}^{↑}.