Identify the critical points. Then use (a) the First...

Identify the critical points. Then use (a) the First Derivative Test and (if possible) (b) the Second Derivative Test to decide which of the critical points give a local maximum and which give a local minimum.
$$f(x)=\frac{x}{x^{2}+4}$$

Key Concepts

Critical Points

Critical points of a function are the values in the domain where its derivative is either zero or undefined. These points are important because changes in the sign of the derivative may indicate transitions between increasing and decreasing behavior, which can correspond to local extrema or saddle points.

First Derivative Test

The First Derivative Test involves analyzing the sign of the derivative before and after each critical point. If the derivative changes from positive to negative, the function has a local maximum at that point; if it changes from negative to positive, the function has a local minimum. This test is useful for understanding the behavior of the function near its critical points.

Second Derivative Test

The Second Derivative Test uses the second derivative to determine the concavity of the function at a critical point. If the second derivative is positive at a critical point, the function is concave up, indicating a local minimum; if it is negative, the function is concave down, indicating a local maximum. If the second derivative is zero, the test is inconclusive, and other methods must be used to classify the point.

Local Maximum

A local maximum is a point in the domain of a function where the function value is greater than all nearby points. In practical terms, around a local maximum, the function increases before the point and decreases after it, which is often confirmed by a positive-to-negative sign change in the derivative.

Local Minimum

A local minimum is a point in the function's domain where the function value is less than all nearby points. Locally, the function decreases before the point and increases after it, corresponding to a negative-to-positive transition in the derivative. This behavior can be confirmed using both the first and second derivative tests.

Transcript

00:01 Now we've got f of x equals x over x squared plus 4.

00:15 I'll use the quotient rule.

00:18 Denominator times the derivative of the numerator, which is just 1, minus numerator times the derivative of the denominator, which is just 2x over denominator squared.

00:42 Simplify that.

00:44 X squared minus 2x squared would be negative x squared.

01:00 Nothing cancels out in the numerator and denominator.

01:04 So in order for this to be 0, then the numerator would have to be 0.

01:12 That would give us x squared equals 4.

01:16 And so x is going to be negative 2 or positive 2.

01:23 Now we'll take the second derivative.

01:27 Again, the equation rule.

01:31 Denominator times the derivative of the numerator, which is negative 2x, minus numerator over derivative, no, no, no, times derivative of the denominator, which would be 2x2x2 times the derivative of the inside, which is 2x.

02:22 Over denominator squared.

02:34 Okay.

02:40 I guess i'll go ahead and simplify this.

02:47 Okay, i see that there's an x squared plus four in every term and in the denominator.

02:54 So i'm going to cancel one out here.

02:58 Cancel that out there.

03:00 And i'm going to call this three.

03:02 Okay...