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Relationship between f, f' and f''

 One of the most important part of calculus is graphing. But for graphing the most important part is to understand the relationship betwwen f, f' and f''. Here is a brief summary of this relationship:






 -root of f (where the function itself crosses the x-axis)

 -the function is always above x-axis

 -the function is always below the x-aixs


 -critical numbers

-possible maximum/minimum  (To confirm, use 1st Derivative Test or 2nd Derivative Test)

 f is increasing 

 f is decreasing                       


 point of inflection

 f is concave upwards (CU)

 f is concave downwards (CD)

 To our common sense, when f is always greater than o, then the function is always above x-axis, and when f is always less than 0, f is always below the x-axis.  And if f is just greater than 0 at certain range, then it is just above x-axis at that corresponding range, vise versa. These have nothing to do with calculus but it is good to know.

 Not hard to discover, when f(0)= 0,  that is the root of the function: when f'(0)=0, then 0 is a critical number and is possible to be max or min. But that is just possible, so in order to confirm, we need to use 1st Derivative Test or 2nd Derivative Test. For 1st Derivative Test, if f' changes from positive to negative at 0, then f has a local max at 0; if f' changes from negative to positive at 0, then f has a local min at 0;if f' does not change signs, 0 is then not local max or min, so you would like to check the two end points for max or min.  For 2nd Derivative Test, suppose f'' is continuous near 0, then if f'(0)=o and f''(o)>0, then f has local min at 0; if f'(o)=0 and f''(0)<0, then f has a local max at 0.      (Here we are just using 0 to discuss, but of course 1st and 2nd Derivative Test can be generalized to any number C.)

 Obviously, when f'>0, f is increasing on the corresponding range; when f'(0)<0, f is decreasing on that range. And when f''>0, f is concave upwards;when f''<0, f is concave downwards.


  • Subject : Math
  • Topic : Calculus
  • Posted By : Jason

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