Math Curve sketching Inflection point

# Inflection point

At an inflection point, the function graph changes its curvature behavior.

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### Remember

#### Necessary criterion

The prerequisite for the presence of inflection points is that the second derivative has a zero at this point:
$f''(x_W)=0$

#### Sufficient criterion

An inflection point exists if in addition:
$f'''(x_W)\neq0$
i

### Method

1. Find derivatives
2. Calculate zero(s) of the second derivative
3. Insert zero(s) in the third derivative
4. Indicate inflection point(s)

### Example

Find the inflection points of the function $f(x)=x^3+2x^2-4x-8$.

1. #### Find derivatives

$f'(x)=3x^2+4x-4$ (the first derivative is not needed)
$f''(x)=6x+4$
$f'''(x)=6$
2. #### Calculate zeros of the second derivative

$f''(x)=6x+4$
$x_W\Leftrightarrow f''(x_W)=0$

$6x+4=0\quad|-4$
$6x=-4\quad|:6$
$x_W=-\frac23$
3. #### Insert zeros into the third derivative

We use the places just identified in the third derivation.
$f'''(x)=6$

$f'''(-\frac23)=6\neq0$
=> There is an inflection point at this position $x=-\frac23$

Note: We use the places just identified in the third derivation.
4. #### Specify inflection points

Inflection points should be specified: Therefore, calculate the y-coordinate with the original function.

$f(-\frac23)$ $=(-\frac23)^3+2\cdot(-\frac23)^2-4\cdot(-\frac23)-8$ $=-4.74$
=> Inflection point: $W(-\frac23|-4.74)$