Math / Algebra

QuestionLet $f(x)=2 x-1$
The smallest positive number for which $f(\cos (f(x)))=0$ is $x=$ $\square$ Hint: Begin by using the definition of $f$ to remove $f$ from the equation. Then solve a linear equation for $\cos (2 x-1)$ . Then solve for $x$ .

Studdy Solution

STEP 1

What is this asking? We need to find the smallest positive $x$ that makes $f(\cos(f(x))) = 0$ , where $f(x) = 2x - 1$ . Watch out! Remember that $\cos(a) = b$ has infinitely many solutions for $a$ , so we'll need to pick the smallest positive one.

STEP 2

1. Substitute and Simplify
2. Solve for Cosine
3. Solve for x
4. Find the Smallest Positive Solution

STEP 3

Let's substitute the definition of $f(x)$ into our equation $f(\cos(f(x))) = 0$ .
Since $f(x) = 2x - 1$ , we have $f(\cos(f(x))) = 2(\cos(f(x))) - 1$ .
So, our equation becomes $2(\cos(f(x))) - 1 = 0$ .

STEP 4

Now, let's substitute $f(x)$ again: $2(\cos(2x - 1)) - 1 = 0$ .
This looks much simpler already!

STEP 5

We want to isolate $\cos(2x - 1)$ .
First, we add 1 to both sides of the equation $2\cos(2x - 1) - 1 + 1 = 0 + 1$ , which simplifies to $2\cos(2x - 1) = 1$ .

STEP 6

Next, we divide both sides by 2: $\frac{2\cos(2x - 1)}{2} = \frac{1}{2}$ , giving us $\cos(2x - 1) = \frac{1}{2}$ .
Great!

STEP 7

We know that $\cos(\frac{\pi}{3}) = \frac{1}{2}$ .
So, one possible solution is $2x - 1 = \frac{\pi}{3}$ .
But remember, cosine is periodic, so we also have $2x - 1 = \frac{\pi}{3} + 2\pi k$ or $2x - 1 = -\frac{\pi}{3} + 2\pi k$ , where $k$ is any integer.

STEP 8

Let's solve for $x$ in the first case: $2x = \frac{\pi}{3} + 1 + 2\pi k$ , so $x = \frac{\pi}{6} + \frac{1}{2} + \pi k$ .

STEP 9

Now, let's solve for $x$ in the second case: $2x = -\frac{\pi}{3} + 1 + 2\pi k$ , so $x = -\frac{\pi}{6} + \frac{1}{2} + \pi k$ .

STEP 10

We want the smallest positive $x$ .
Let's try $k = 0$ in the first case: $x = \frac{\pi}{6} + \frac{1}{2} \approx 0.52 + 0.5 = 1.02$ .
This is positive!

STEP 11

Now let's try $k = 0$ in the second case: $x = -\frac{\pi}{6} + \frac{1}{2} \approx -0.52 + 0.5 = -0.02$ .
This is negative, so we need to try $k = 1$ : $x = -\frac{\pi}{6} + \frac{1}{2} + \pi \approx -0.52 + 0.5 + 3.14 = 3.12$ .
This is positive, but larger than our first solution.

STEP 12

Therefore, the smallest positive solution is $x = \frac{\pi}{6} + \frac{1}{2}$ .

STEP 13

The smallest positive number for which $f(\cos(f(x))) = 0$ is $x = \frac{\pi}{6} + \frac{1}{2}$ .

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QuestionLet $f(x)=2 x-1$
The smallest positive number for which $f(\cos (f(x)))=0$ is $x=$ $\square$ Hint: Begin by using the definition of $f$ to remove $f$ from the equation. Then solve a linear equation for $\cos (2 x-1)$ . Then solve for $x$ .

Studdy Solution

STEP 1

What is this asking? We need to find the smallest positive $x$ that makes $f(\cos(f(x))) = 0$ , where $f(x) = 2x - 1$ . Watch out! Remember that $\cos(a) = b$ has infinitely many solutions for $a$ , so we'll need to pick the smallest positive one.

STEP 2

1. Substitute and Simplify
2. Solve for Cosine
3. Solve for x
4. Find the Smallest Positive Solution

STEP 3

Let's substitute the definition of $f(x)$ into our equation $f(\cos(f(x))) = 0$ .
Since $f(x) = 2x - 1$ , we have $f(\cos(f(x))) = 2(\cos(f(x))) - 1$ .
So, our equation becomes $2(\cos(f(x))) - 1 = 0$ .

STEP 4

Now, let's substitute $f(x)$ again: $2(\cos(2x - 1)) - 1 = 0$ .
This looks much simpler already!

STEP 5

We want to isolate $\cos(2x - 1)$ .
First, we add 1 to both sides of the equation $2\cos(2x - 1) - 1 + 1 = 0 + 1$ , which simplifies to $2\cos(2x - 1) = 1$ .

STEP 6

Next, we divide both sides by 2: $\frac{2\cos(2x - 1)}{2} = \frac{1}{2}$ , giving us $\cos(2x - 1) = \frac{1}{2}$ .
Great!

STEP 7

We know that $\cos(\frac{\pi}{3}) = \frac{1}{2}$ .
So, one possible solution is $2x - 1 = \frac{\pi}{3}$ .
But remember, cosine is periodic, so we also have $2x - 1 = \frac{\pi}{3} + 2\pi k$ or $2x - 1 = -\frac{\pi}{3} + 2\pi k$ , where $k$ is any integer.

STEP 8

Let's solve for $x$ in the first case: $2x = \frac{\pi}{3} + 1 + 2\pi k$ , so $x = \frac{\pi}{6} + \frac{1}{2} + \pi k$ .

STEP 9

Now, let's solve for $x$ in the second case: $2x = -\frac{\pi}{3} + 1 + 2\pi k$ , so $x = -\frac{\pi}{6} + \frac{1}{2} + \pi k$ .

STEP 10

We want the smallest positive $x$ .
Let's try $k = 0$ in the first case: $x = \frac{\pi}{6} + \frac{1}{2} \approx 0.52 + 0.5 = 1.02$ .
This is positive!

STEP 11

Now let's try $k = 0$ in the second case: $x = -\frac{\pi}{6} + \frac{1}{2} \approx -0.52 + 0.5 = -0.02$ .
This is negative, so we need to try $k = 1$ : $x = -\frac{\pi}{6} + \frac{1}{2} + \pi \approx -0.52 + 0.5 + 3.14 = 3.12$ .
This is positive, but larger than our first solution.

STEP 12

Therefore, the smallest positive solution is $x = \frac{\pi}{6} + \frac{1}{2}$ .

STEP 13

The smallest positive number for which $f(\cos(f(x))) = 0$ is $x = \frac{\pi}{6} + \frac{1}{2}$ .

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Studdy Solution

STEP 1

STEP 2

1. Substitute and Simplify2. Solve for Cosine3. Solve for x4. Find the Smallest Positive Solution

STEP 3

STEP 4

Now, let's **substitute** f(x)f(x)f(x) again: 2(cos⁡(2x−1))−1=02(\cos(2x - 1)) - 1 = 02(cos(2x−1))−1=0.This looks much simpler already!

STEP 5

We want to **isolate** cos⁡(2x−1)\cos(2x - 1)cos(2x−1).First, we **add** 1 to both sides of the equation 2cos⁡(2x−1)−1+1=0+12\cos(2x - 1) - 1 + 1 = 0 + 12cos(2x−1)−1+1=0+1, which simplifies to 2cos⁡(2x−1)=12\cos(2x - 1) = 12cos(2x−1)=1.

STEP 6

Next, we **divide** both sides by 2: 2cos⁡(2x−1)2=12\frac{2\cos(2x - 1)}{2} = \frac{1}{2}22cos(2x−1)​=21​, giving us cos⁡(2x−1)=12\cos(2x - 1) = \frac{1}{2}cos(2x−1)=21​.Great!

STEP 7

STEP 8

Let's **solve** for xxx in the first case: 2x=π3+1+2πk2x = \frac{\pi}{3} + 1 + 2\pi k2x=3π​+1+2πk, so x=π6+12+πkx = \frac{\pi}{6} + \frac{1}{2} + \pi kx=6π​+21​+πk.

STEP 9

Now, let's **solve** for xxx in the second case: 2x=−π3+1+2πk2x = -\frac{\pi}{3} + 1 + 2\pi k2x=−3π​+1+2πk, so x=−π6+12+πkx = -\frac{\pi}{6} + \frac{1}{2} + \pi kx=−6π​+21​+πk.

STEP 10

We want the **smallest positive** xxx.Let's try k=0k = 0k=0 in the first case: x=π6+12≈0.52+0.5=1.02x = \frac{\pi}{6} + \frac{1}{2} \approx 0.52 + 0.5 = 1.02x=6π​+21​≈0.52+0.5=1.02.This is positive!

STEP 11

STEP 12

Therefore, the **smallest positive** solution is x=π6+12x = \frac{\pi}{6} + \frac{1}{2}x=6π​+21​.

STEP 13

The smallest positive number for which f(cos⁡(f(x)))=0f(\cos(f(x))) = 0f(cos(f(x)))=0 is x=π6+12x = \frac{\pi}{6} + \frac{1}{2}x=6π​+21​.

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1. Substitute and Simplify
2. Solve for Cosine
3. Solve for x
4. Find the Smallest Positive Solution

Now, let's substitute $f(x)$ again: $2(\cos(2x - 1)) - 1 = 0$ .
This looks much simpler already!

We want to isolate $\cos(2x - 1)$ .
First, we add 1 to both sides of the equation $2\cos(2x - 1) - 1 + 1 = 0 + 1$ , which simplifies to $2\cos(2x - 1) = 1$ .

Next, we divide both sides by 2: $\frac{2\cos(2x - 1)}{2} = \frac{1}{2}$ , giving us $\cos(2x - 1) = \frac{1}{2}$ .
Great!

Let's solve for $x$ in the first case: $2x = \frac{\pi}{3} + 1 + 2\pi k$ , so $x = \frac{\pi}{6} + \frac{1}{2} + \pi k$ .

Now, let's solve for $x$ in the second case: $2x = -\frac{\pi}{3} + 1 + 2\pi k$ , so $x = -\frac{\pi}{6} + \frac{1}{2} + \pi k$ .

We want the smallest positive $x$ .
Let's try $k = 0$ in the first case: $x = \frac{\pi}{6} + \frac{1}{2} \approx 0.52 + 0.5 = 1.02$ .
This is positive!

Therefore, the smallest positive solution is $x = \frac{\pi}{6} + \frac{1}{2}$ .

The smallest positive number for which $f(\cos(f(x))) = 0$ is $x = \frac{\pi}{6} + \frac{1}{2}$ .