Math / Calculus

QuestionLet $g(x)=\sin (2 x)$ , for $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ . Where does $g$ have critical points?

Studdy Solution

STEP 1

What is this asking? Where does the sine wave $g(x) = \sin(2x)$ flatten out between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ ? Watch out! Remember that critical points can be minimums, maximums or places where the derivative doesn't exist, but in this case, our function is nice and smooth!

STEP 2

1. Find the derivative
2. Set the derivative to zero
3. Solve for x

STEP 3

Alright, let's kick things off by finding the derivative of our function $g(x) = \sin(2x)$ .
We'll use the chain rule here.
Remember, the chain rule helps us find the derivative of composite functions, like our sine wave.

STEP 4

The derivative of the outer function $\sin(u)$ is $\cos(u)$ .
The derivative of the inner function $u = 2x$ is just $2$ .

STEP 5

Multiplying these together gives us our derivative: $g'(x) = 2 \cdot \cos(2x)$ .
Awesome!

STEP 6

Now, we want to find where this derivative is equal to zero.
This is where our function $g(x)$ will have its critical points.
So, we set $g'(x) = 0$ , which gives us $2 \cdot \cos(2x) = 0$ .

STEP 7

Let's isolate the cosine term by dividing both sides of the equation by $2$ .
This gives us $\cos(2x) = 0$ .
Remember, dividing both sides by $2$ is like multiplying both sides by $\frac{1}{2}$ , and $\frac{1}{2} \cdot 2 = 1$ , which is why the $2$ disappears from the left side.

STEP 8

Now, we need to think: when does the cosine function equal zero?
Cosine represents the x-coordinate on the unit circle.
It's zero at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ , and also at their negative counterparts.

STEP 9

So, we can say $2x = \frac{\pi}{2}$ or $2x = \frac{3\pi}{2}$ or $2x = -\frac{\pi}{2}$ or $2x = -\frac{3\pi}{2}$ .

STEP 10

To solve for $x$ , we'll divide each of these by $2$ .
This gives us $x = \frac{\pi}{4}$ , $x = \frac{3\pi}{4}$ , $x = -\frac{\pi}{4}$ , and $x = -\frac{3\pi}{4}$ .

STEP 11

But wait!
Remember our initial interval for $x$ ?
It was $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ .
So, $\frac{3\pi}{4}$ and $-\frac{3\pi}{4}$ are outside of this range.
Our final critical points are $x = \frac{\pi}{4}$ and $x = -\frac{\pi}{4}$ .
Woohoo!

STEP 12

The critical points of $g(x) = \sin(2x)$ on the interval $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ are $x = \frac{\pi}{4}$ and $x = -\frac{\pi}{4}$ .

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms