Difference between revisions of "First derivative test"

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(Notes)
(Notes)
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==Notes==
 
==Notes==
  
===Points of inflection===
+
===First derivative test does not require differentiability at the point===
  
===Examples illustrating why the test is not always conclusive===
+
To apply the two-sided combined sign version of the first derivative test, we need ''continuity'' at the point and differentiability on the immediate left and immediate right of the point. However, we do not require differentiability ''at'' the point.
  
The following problems could occur when applying this test:
+
Thus, for instance, the first derivative test can be used to study the behavior of a function with a piecewise definition by interval, such that the function is changing definition at the point. Explicitly, it can be used to study functions of the form:
  
# The function is not continuous, or not differentiable, at points to the immediate left or immediate right of the critical point.
+
<math>f(x) := \lbrace \begin{array}{rl} f_1(x), & x < c \\ f_2(x), & c < x \\ v, & x = c \\\end{array}</math>
# The function is differentiable on the immediate left and immediate right of the critical point. However, the derivative does not have a uniform sign on the immediate left or the immediate right, i.e., it is oscillatory in sign at points arbitrarily close to the critical point.
 
  
Here is a picture of a function illustrating (2):
+
Assume that <math>f</math> is continuous at <math>c</math>, i.e., <math>\lim_{x \to c^-} f_1(x) = \lim_{x \to c^+} f_2(x) = v</math>. In that case, we can try to determine whether <math>c</math> is a point of local maximum, minimum, or neither by studying the sign of <math>f_1'</math> to the immediate left of <math>c</math> and the sign of <math>f_2'</math> to the immediate right of <math>c</math>. It is not necessary that <math>f</math> be differentiable at <math>c</math> (for more on how to differentiate piecewise functions, see [[differentiation rule for piecewise definition by interval]]).
  
[[File:Firstderivativetestfails.png|500px]]
+
===Examples illustrating why the test is not always conclusive===
  
The function illustrated in the picture is:
+
The following problems could occur when applying this test:
  
<math>f(x) := \lbrace\begin{array}{rl} |x|\left(2 + \sin\left(\frac{1}{x}\right)\right), & x \ne 0 \\ 0, & x = 0 \\\end{array}</math>
+
# [[First derivative test fails for function that is discontinuous at the critical point]]: If the function is not continuous at the critical point, then the first derivative test may yield incorrect conclusions.
 +
# The first derivative test fails (or rather, cannot be applied) if the function is not differentiable on the immediate left or immediate right of the point.
 +
# [[First derivative test fails for function whose derivative has ambiguous sign around the point]]: A pictorial illustration is below:
  
We note the following:
+
[[File:Firstderivativetestfails.png|500px]]
 
 
* <math>f</math> has a local and absolute minimum at 0: For <math>x \ne 0</math>, we have <math>f(x) > 0</math>. This is because <math>|x| > 0</math> and <math>2 + \sin(1/x) \in [1,3]</math>, so that is also positive.
 
* <math>f</math> is continuous at 0: We can see this using the [[pinching theorem]] or more directly by noting that as <math>x \to 0</math>, <math>|x| \to 0</math> and <math>2 + \sin(1/x)</math> is bounded between finite values 1 and 3.
 
* <math>f</math> has ambiguous sign on the immediate right of 0: For <math>x > 0</math>, we have <math>f(x) = x(2 + \sin(1/x))</math>. The derivative is <math>f'(x) = 2 + \sin(1/x) - (1/x)\cos(1/x)</math>. The part <math>2 + \sin(1/x)</math> is bounded, but the part <math>-(1/x)\cos(1/x)</math> oscillates between large magnitude positive and negative values as <math>x \to 0^+</math>. In particular, <math>f'</math> does not have constant sign on the immediate right of 0.
 
* <math>f</math> has ambiguous sign on the immediate left of 0: For <math>x < 0</math>, we have <math>f(x) = -x(2 + \sin(1/x))</math>. The derivative is <math>f'(x) = -2 - \sin(1/x) + (1/x)\cos(1/x)</math>. The part <math>-2 - \sin(1/x)</math> is boundd, but the part <math>(1/x)\cos(1/x)</math> oscillates between large magnitude positive and negative values as <math>x \to 0^-</math>. In particular, <math>f'</math> does not have constant sign on the immediate left of 0.
 

Revision as of 16:41, 24 April 2012

Statement

What the test is for

The first derivative test is a partial (i.e., not always conclusive) test used to determine whether a particular critical point in the domain of a function is a point where the function attains a local maximum value, local minimum value, or neither. There are cases where the test is inconclusive, which means that we cannot draw any conclusion.

What the test says: one-sided sign versions

Suppose f is a function defined at a point c.

Then, we have the following:

Continuity and differentiability assumption Hypothesis on sign of derivative Conclusion
f is left continuous at c and differentiable on the immediate left of c \! f'(x) is positive (respectively, nonnegative) for x to the immediate left of c (i.e., for x \in (c - \delta, c) for sufficiently small \delta > 0) f has a strict local maximum from the left at c, i.e., f(c) > f(x) (respectively, f has a local maximum from the left at c, i.e., f(c) \ge f(x)) for x to the immediate left of c.
f is left continuous at c and differentiable on the immediate left of c \! f'(x) is negative (respectively, nonpositive) for x to the immediate left of c (i.e., for x \in (c - \delta, c) for sufficiently small \delta > 0) f has a strict local minimum from the left at c, i.e., f(c) < f(x) (respectively, f has a local minimum from the left at c, i.e., f(c) \le f(x)) for x to the immediate left of c.
f is right continuous at c and differentiable on the immediate right of c \! f'(x) is positive (respectively, nonnegative) 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 minimum from the right at c, i.e., f(c) < f(x) (respectively, f has a local minimum from the right at c, i.e., f(c) \le f(x)) for x to the immediate right of c.
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.

What the test says: combined sign versions

Suppose f is a function defined around a point c (i.e., f is defined in an open interval containing c) and is continuous at c. We do not care whether f is differentiable at c; however, the test makes sense only if f is differentiable on the immediate left and immediate right of c.

Then, we have the following (we list only the strict cases in the table below):

Continuity and differentiability assumption Sign of the derivative f' on immediate left of c Sign of f' on immediate right of c Conclusion about local minimum, local maximum, or neither
f is continuous at c and differentiable on the immediate left and immediate right of c positive negative strict local maximum
f is continuous at c and differentiable on the immediate left and immediate right of c negative positive strict local minimum
f is continuous at c and differentiable on the immediate left and immediate right of c positive positive neither local maximum nor local minimum, because f is increasing through the point
f is continuous at c and differentiable on the immediate left and immediate right of c negative negative neither local maximum nor local minimum, because f is decreasing through the point

If we replace positive by nonnegative and negative by nonpositive in the rows corresponding to strict local maximum and strict local minimum, we could potentially lose the strictness.

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

The typical goal of the first derivative test is to determine whether a critical point is a point of local maximum or minimum. Hence, the test is typically applied to critical points. However, when applying the first derivative test, we do not need to check whether the point in question is a critical point. In other words, if the condition for being a point of local maximum or minimum is satisfied, then the point in question is automatically a critical point and this condition need not be checked separately.

Succinct version

Here is a shorter version: at a critical point, if the derivative changes sign from negative to positive (as we go from left to right) then that is a point of local minimum. If the derivative changes sign from positive to negative (as we go from left to right) then it is a point of local maximum.

Related tests

Notes

First derivative test does not require differentiability at the point

To apply the two-sided combined sign version of the first derivative test, we need continuity at the point and differentiability on the immediate left and immediate right of the point. However, we do not require differentiability at the point.

Thus, for instance, the first derivative test can be used to study the behavior of a function with a piecewise definition by interval, such that the function is changing definition at the point. Explicitly, it can be used to study functions of the form:

f(x) := \lbrace \begin{array}{rl} f_1(x), & x < c \\ f_2(x), & c < x \\ v, & x = c \\\end{array}

Assume that f is continuous at c, i.e., \lim_{x \to c^-} f_1(x) = \lim_{x \to c^+} f_2(x) = v. In that case, we can try to determine whether c is a point of local maximum, minimum, or neither by studying the sign of f_1' to the immediate left of c and the sign of f_2' to the immediate right of c. It is not necessary that f be differentiable at c (for more on how to differentiate piecewise functions, see differentiation rule for piecewise definition by interval).

Examples illustrating why the test is not always conclusive

The following problems could occur when applying this test:

  1. First derivative test fails for function that is discontinuous at the critical point: If the function is not continuous at the critical point, then the first derivative test may yield incorrect conclusions.
  2. The first derivative test fails (or rather, cannot be applied) if the function is not differentiable on the immediate left or immediate right of the point.
  3. First derivative test fails for function whose derivative has ambiguous sign around the point: A pictorial illustration is below:

Firstderivativetestfails.png