Differentiation: Applications of derivatives
Inflection points
An inflection point, or point of inflection, is a point where the graph of a function makes a kind of "bend" by changing the type of increase or decrease. This means that the derivative of the function changes from increasing to decreasing, or vice versa, and thus has a local maximum or minimum. These points can only occur if the second derivative is #0#.
An inflection point of a graph is a point in which the graph changes its type of increase or decrease. The graph can change:
 from #\blue{\text{concave up and increasing}}\ (long\_short\ dashed)# to #\green{\text{concave down and increasing}}\ (dashed)#
 from #\green{\text{concave down and increasing}}\ (dashed)# to #\blue{\text{concave up and increasing}}\ (long\_short\ dashed)#
 from #\orange{\text{concave down and decreasing}}\ (dotted)# to #\purple{\text{concave up and decreasing}}\ (solid)#
 from #\purple{\text{concave up and decreasing}}\ (solid)# to #\orange{\text{concave down and decreasing}}\ (dotted)#
In the image to the right, #x=2# is an inflection point.
We can find the inflection points using the second derivative.
A function #\blue{f(x)}# has an inflection point in a point #\orange{c}# if #f'(\orange{c})# is a local minimum or maximum. The following then applies:
\[\green{f''(\orange{c})}=0\]
Example
\[\begin{array}{rcl} \blue{f(x)}&=&\blue{x^3}\\ f'(x)&=&3x^2\\ \green{f''(x)}&=&\green{6x}\\ \green{f''(\orange{0})}&=&0\end{array}\]
Stepbystep Calculating inflection points 
Example 

Determine the inflection points for a function #\blue{f(x)}#. 
#\begin{array}{rcl}\blue{f(x)}&=&\phantom{'}\blue{x^5x^3}\end{array}# 

Step 1 
Calculate the first derivative #f'(x)#. 
#\begin{array}{rcl}f'(x)&=&5x^43x^2\end{array}# 
Step 2 
Calculate the second derivative #\green{f''(x)}#. 
#\begin{array}{rcl}\green{f''(x)}&=&\green{20x^36x}\end{array}# 
Step 3 
Solve #f''(x)=0# to find the possible local minima and maxima of #f'(x)# and with them the inflection points of #f(x)#. 
#\begin{array}{c}{x}={0} \lor {x}={\sqrt{\tfrac{3}{10}} }\;\lor\; {x}={\sqrt{\tfrac{3}{10}}}\end{array}# 
Step 4 
Determine whether the values found in step 3 belong to a local minimum or maximum. If so, then it is an inflection point. 
#{x}={\sqrt{\tfrac{3}{10}}}# local minimum #f'# #{x}={0}# local maximum #f'# #{x}={\sqrt{\tfrac{3}{10}}}# local minimum #f'# Hence all three inflection points #f(x)# 
Give your answer in the form #x=x_1 \lor x=x_2 \lor \ldots\lor x=x_n# if there are #n# number of inflection points, in the form #x=x_1# if there is one point of inflection, and in the form #none# if there are no inflection points.
Step 1  We determine the derivative of #f(x)=x^9# using the power rule. This is equal to: \[f'(x)=9\cdot x^8\] 
Step 2  We determine the second derivative in the same way. This is equal to: \[f''(x)=72\cdot x^7\] 
Step 3  We solve the equation #72\cdot x^7=0#. This goes as follows: \[\begin{array}{rcl} 72\cdot x^7&=&0\\&&\phantom{xxx}\blue{\text{the equation we need to solve}}\\ x^7&=&0\\&&\phantom{xxx}\blue{\text{both sides divided by }72}\\ x&=&0\\&&\phantom{xxx}\blue{\text{got rid of the exponents on both sides}}\end{array} \] 
Step 4  We draw the graph of #f'(x)=9\cdot x^8#. We see that #f'(x)# has a local minimum at #x=0#. Therefore, the function #f(x)# has an inflection point at #x=0#. 
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