Determining Points of Inflection for a Function
Add yours
Key Questions

Make the second derivative equation equal to 0, and find the x values.
When the second derivative is above 0, the original function is con caved up. If it is below 0, the original function is concave down.

#f(x)=x^3+x# By taking derivatives,
#f'(x)=3x^2+1# #f''(x)=6x=0 Rightarrow x=0# ,which is the
#x# coordinate of a possible inflection point. (We still need to verify that#f# changes its concavity there.)Use
#x=0# to split#(infty,\infty)# into#(infty,0)# and#(0,infty)# .Let us check the signs of
#f''# at sample points#x=1# and#x=1# for the intervals, respectively.
(You may use any number on those intervals as sample points.)#f''(1)=6<0 Rightarrow f# is concave downward on#(infty,0)# #f''(1)=6>0 Rightarrow f# is concave upward on#(0,infty)# Since the above indicates that
#f# changes its concavity at#x=0# ,#(0,f(0))=(0,0)# is an inflection point of#f# .I hope that this was helpful.

This key question hasn't been answered yet. Answer question
Questions
























Doublecheck the answer













Videos on topic View all (2)
Graphing with the Second Derivative

1Relationship between First and Second Derivatives of a Function

2Analyzing Concavity of a Function

3Notation for the Second Derivative

4Determining Points of Inflection for a Function

5First Derivative Test vs Second Derivative Test for Local Extrema

6The special case of x⁴

7Critical Points of Inflection

8Application of the Second Derivative (Acceleration)

9Examples of Curve Sketching