Find the value(s) of x for which: (a) f(x) has a local...

Key Concepts

Differentiation of Trigonometric Functions

Understanding the differentiation of trigonometric functions is crucial since many functions involve sine, cosine, or other trigonometric forms. Mastery of these derivative rules allows for analyzing the behavior of such functions, identifying critical points, and effectively applying derivative tests.

Extreme Value Theorem and Closed Interval Method

The Extreme Value Theorem guarantees that a continuous function defined on a closed interval will have both a global maximum and a global minimum. This concept is fundamental in optimization problems, as it necessitates evaluating the function at both the critical points and the interval's endpoints to ensure that all possible extrema are identified.

First and Second Derivative Tests

The first derivative test analyzes the change in the sign of the derivative around a critical point to determine whether the function experiences a local maximum or minimum there. The second derivative test further aids by evaluating the concavity of the function at critical points, with positive second derivatives indicating a local minimum and negative ones indicating a local maximum.

Critical Points

Critical points are values in the domain of a function where its derivative is zero or undefined, and they represent potential locations for local extrema. Identifying critical points is essential in the study of calculus as they help pinpoint where a function may change direction or behavior.

Transcript

00:01 We have f of x is equal to sine squared x minus cosine x.

00:10 And we wish to find the values of x for which this function has a local max or a local min.

00:16 So the interval that we're given is x is between zero and pi.

00:24 So now part a, to find local max and local min, you'll have to find the critical points.

00:30 And to find critical points, you just need to take the derivative.

00:33 Set it equal to zero, solve for x.

00:37 Or look for the values of x where f prime of x doesn't exist.

00:41 So let's start by finding f prime of x.

00:46 So this is going to be equal to 2 sine x.

00:51 And then chain rule says it'll be cosine x.

00:54 And then minus the derivative of cosine is negative sign.

00:58 So this is plus sinex.

01:00 So let's factor this.

01:02 This is sine x times 2 cos x plus 1.

01:10 So now sine x and cosine x are both continuous functions everywhere.

01:17 So this means that f prime of x exists everywhere.

01:20 So the only critical values that we will find would be when we set f prime of x equal to zero.

01:27 So f prime of x is equal to zero, equal to sine x times two coase x plus one, which means either sine x is equal to zero or two cos x plus one is equal to zero.

01:49 So sine x is equal to zero at zero and pi.

01:53 Well, two pi as well and so on, but we only care about the values within our interval.

01:59 So from sine x is equal to zero, we get x is equal to 0 and pi.

02:07 And then from 2, cos x plus 1 is equal to 0, we should get cos x is equal to negative 1 over 2.

02:17 Okay, so you would use your special triangle for this.

02:20 So when is cosine x equal to negative 1 over 2? well, the reference angle for that should be pi over 3.

02:31 Okay, but we need cosine to be negative, which means x has to be equal to.

02:39 If you draw your, let's draw that on the side.

02:43 I draw a small one.

02:45 C -a -s -t.

02:49 And the reference angle is pi over 3, and we only care about the interval from 0 to pi, which means from here to here.

02:59 So the only one that we can draw is this one here, with the reference angle of pi over 3.

03:05 So which means x is going to, to be equal to 2 pi over 2.

03:13 Okay, so now we found our critical points.

03:16 Now we need to classify them either as a local max or a local myth.

03:21 And to do that, you can use your second derivative test.

03:26 So if i do my second derivative test, i get f double prime is equal to.

03:32 So take the derivative of that.

03:34 That's going to be 2 .2 squared x plus, or that should be actually a minus, 2 sine squared x, and then plus cosine x.

03:54 Okay, so here you can use your identity.

03:57 If you don't remember, that's fine.

04:00 We'll factor out the two, and then we can actually replace sine squared x with one minus co -squared x.

04:08 So this is really co -squared x minus one plus co -squared x.

04:15 And this is plus cos x...

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