Are the inflection points where f'(x) = zero or where the graph changes from concave up to concave down?

2 Answers
May 23, 2015

The inflection point is a point where the graph of the function changes from concave up to concave down or vice versa.

To calculate these points you have to find places where #f''(x)=0# and check if the second derivative changes sign at this point.

For example to find the points of inflection for #f(x)=x^7#you have to calculate #f''(x)# first.

#f'(x)=7x^6#
#f''(x)=42x^5#

Now we have to check where #f''(x)=0#
#42x^5=0 iff x=0#.

We found that #x=0# may be a point of inflection.
To find if it is such point we have to check if #f''(x)# changes sign at 0.

To find this we can graph the function:

graph{42x^5 [-3.894, 3.897, -1.95, 1.948]}

We can see that the #f''(x)# changes sign at zero, so zero is the point of inflection.

Note
It is important to check to see whether concavity actually changes.

#g(x)=x^4+3x-8#

#g'(x)+ 4x^3+3#

#g''(x) =12x^2#

Now we have to check where #g''(x)=0#
#12x^2=0 iff x=0#.

We found that #x=0# may be a point of inflection.
To find if it is such point we have to check if #g''(x)# changes sign at #0#.

But #g''(x) =12x^2# is never negative, it is always positive or #0#, therefors the sign of #g''(x)# does not change, so there are no inflection points.

May 23, 2015

Answer:

I have been taught and, following our textbook's lead, I continue to teach , that an inflection point is a point on the graph at which the concavity changes.

Explanation:

Using this terminology:
For #f(x) = x^7#, the inflection point is #(0,0)#

The function: #h(x) = 1/x# is concave down on #(-oo,0)# and concave up on #(0,oo)#.
The concavity is not the same on the entire graph, but there is no inflection point, because there is no point on the graph at #x=0#.