# Changes

## First derivative test

, 21:14, 7 March 2013
What the test says: one-sided sign versions
{{maxmin test}}
==Statement==
| $f$ is right continuous ''at'' $c$ and differentiable on the immediate right of $c$ ||$\! f'(x)$ is negative (respectively, nonpositive) for $x$ to the immediate right of $c$ (i.e., for $x \in (c,c + \delta)$ for sufficiently small $\delta > 0$)|| $f$ has a strict local maximum from the right at $c$, i.e., $f(c) > f(x)$ (respectively, $f$ has a local maximum from the right at $c$, i.e., $f(c) \ge f(x)$) for $x$ to the immediate right of $c$. || [[File:Rightdecreasingconcavedownnotflat.png|100px]][[File:Rightdecreasingconcaveup.png|80px]]
|}

===What the test says: combined sign versions===
Note that if $f'$ has ambiguous sign on the immediate left or on the immediate right of $c$, the first derivative test is inconclusive.

===Relation with critical points===
* In general, if the derivative changes sign as we move from the immediate left of the point to the immediate right of the point, then there is a local extremum at the point. If the derivative has the same sign on the immediate left and immediate right, we ''do not'' get a local extremum at the point.
==Facts used==
|}
<center>{{#widget:YouTube|id=6PJplELQB1g}}</center> =When is the test =Inconclusive and conclusive and inconclusive?cases==
===Situations when the test is inconclusiveInconclusive cases===
Note that we consider the first derivative test to be conclusive if we can definitely conclude whether we have a local maximum, local minimum, or neither. In particular, the first derivative test is conclusive for a function that's continuous at the point, differentiable on the immediate left and immediate right of the point, and whose derivative takes constant sign (possibly allowing zero values) on the immediate left and constant sign (possibly allowing zero values) on the immediate right.
{| class="sortable" border="1"
! What problem do we run into? !! What kind of trouble can we have? !! Link to example !! Can this be fixed? Remedy that may work !! Picture
|-
| The function is not continuous at the critical point || We may be able to do sign analysis of the derivative on the immediate left and immediate right, but draw incorrect conclusions by applying the one-sided or combined sign version of the first derivative test. A priori, all the possibilities (local maximum, local minimum, neither) remain open. || [[first derivative test fails for function that is discontinuous at the critical point]] || If the function has one-sided limits at the critical point: [[variation of first derivative test for discontinuous function with one-sided limits]] ||
|}
<center>{{#widget:YouTube|id=h-pgA5s_CEY}}</center> ==Condition for the test to be conclusive=Conclusive cases===
* [[First derivative test is conclusive for differentiable function at isolated critical point]]: If $f$ is continuous at $c$ and differentiable on the immediate left and immediate right of a [[critical point]] $c$, ''and'' $c$ is an isolated critical point (i.e., there is an open interval containing it that contains no other critical points), then the first derivative test must be conclusive at $c$. In other words, we can use the first derivative test to definitively determine whether $c$ is a point of local maximum, local minimum, or neither, for $f$.
* [[First derivative test is conclusive for function with algebraic derivative]]: This includes polynomials and rational functions.
* [[First derivative test is conclusive for locally analytic function]]
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