Difference between revisions of "First derivative test"

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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 positive to negative (as we go from left to right) then it is a point of local maximum. If the derivative changes sign from negative to positive (as we go from left to right) then that is a point of local minimum.

Proof

Example proof of one-sided version: positive derivative on left

All the one-sided versions have analogous proofs, so we provide a proof only for one of them.

Given: A function $f$ and a point $c$ in the domain. $f$ is left continuous at $c$ and differentiable on the immediate left of $c$. Further, $f'(x) > 0$ on the immediate left of $c$. Explicitly, there exists $\delta > 0$ such that $f'(x) > 0$ for $x \in (c - \delta, c)$.

To prove: $f$ has a strict local maximum from the left at $c$. More explicitly, we have $f(x) < f(c)$ for $x \in (c - \delta, c)$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $f$ is increasing on the immediate left of $c$, i.e., $f$ is increasing on the interval $(c - \delta, c)$. Fact (1) $f'(x) > 0$ for $x \in (c - \delta, c)$ given-fact direct
2 $f$ is increasing from the left up to and including $c$, i.e., $f$ is increasing on $(c - \delta,c]$. Fact (2) $f$ is left continuous at $c$ Step (1) step-given-fact direct
3 $f(x) < f(c)$ for $x \in (c - \delta, c)$ Step (2) Follows directly from Step (2).

Example proof of combined sign version: strict local maximum

We give the proof for the strict local maximum case. Other cases are analogous.

Given: A function $f$ and a point $c$ in the domain. $f$ is continuous at $c$ and differentiable on the immediate left and immediate right of $c$. Further, $f'(x) > 0$ on the immediate left of $c$ and $f'(x) < 0$ on the immediate right of $c$.

To prove: $f$ has a two-sided strict local maximum at $c$, i.e., $f(x) < f(c)$ for $x$ on the immediate left or the immediate right of $c$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $f$ has a strict local maximum from the left at $c$ one-sided version for strict local max from the left $f$ is continuous at $c$ and $f'(x) > 0$ on the immediate left of $c$ Since $f$ is continuous at $c$, it is in particular left continuous at $c$. Combining this with $f'(x) > 0$ (on the immediate left) and the one-sided sign version of the first derivative test, we obtain the result.
2 $f$ has a strict local maximum from the right at $c$ one-sided version for strict local max from the right $f$ is continuous at $c$ and $f'(x) < 0$ on the immediate right of $c$ Since $f$ is continuous at $c$, it is in particular right continuous at $c$. Combining this with $f'(x) < 0$ (on the right of $c$) and the one-sided sign version of the first derivative test, we obtain the result.
3 $f$ has a two-sided strict local maximum at $c$ Steps (1), (2) Step-combination direct

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).

When is the test conclusive and inconclusive?

Situations when the test is inconclusive

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.

The following problems could occur when applying this test:

What problem do we run into? What kind of trouble can we have? Link to example Can this be fixed? 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 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
The function is not differentiable at points on the immediate left and/or immediate right of the point We will not be able to make a meaningful statement about the sign of the derivative on the immediate left and/or immediate right. Thus, it will not be possible to apply the first derivative test. All the possibilities (local maximum, local minimum, neither) remain open. -- Not directly. We have to use other methods.
The derivative of the function has oscillatory (ambiguous) sign on the immediate left and/or immediate right of the point We cannot do sign analysis on the derivative on the immediate left and/or immediate right. Thus, it will not be possible to apply the first derivative test. All the possibilities (local maximum, local minimum, neither) remain open. First derivative test is inconclusive for function whose derivative has ambiguous sign around the point

Condition for the test to be conclusive

• 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 $f$.
• In particular, the first derivative test is always conclusive for polynomials and rational functions. It is also conclusive for functions with piecewise definition by interval where each of the piece is a polynomial or rational function.