Nilai F(11) Jika F(3x + 2) = 18x + 3

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Hey guys! Let's dive into this math problem together. We're given the function f(3x + 2) = 18x + 3, and our mission is to find the value of f(11). Sounds like fun, right? Let's break it down step by step.

Understanding the Problem

First, we need to figure out what value of x will make the expression inside the function, 3x + 2, equal to 11. In other words, we want to solve the equation:

3x + 2 = 11

Once we find that x, we can plug it into the right side of the original equation, 18x + 3, to find the value of f(11). Simple enough, right?

Solving for x

Let's solve for x:

3x + 2 = 11

Subtract 2 from both sides:

3x = 9

Now, divide both sides by 3:

x = 3

Alright! We found that x = 3.

Finding f(11)

Now that we know x = 3, we can plug this value into the expression 18x + 3:

f(11) = 18(3) + 3

f(11) = 54 + 3

f(11) = 57

So, the value of f(11) is 57. Looking at the options, that matches option D. Awesome!

Conclusion

Therefore, if f(3x + 2) = 18x + 3, then the value of f(11) is indeed 57. You did it! This problem demonstrates a common type of function evaluation question, where you need to manipulate the input of the function to match the desired value. Keep practicing, and you'll become a pro at these in no time! Remember, the key is to break down the problem into smaller, manageable steps.

Let's recap the steps we took:

  1. Set the expression inside the function equal to the desired value: 3x + 2 = 11
  2. Solve for x.
  3. Substitute the value of x into the other side of the equation to find the value of the function at the desired input.

With these steps, you'll be able to tackle similar problems with confidence. Great job, and keep up the excellent work!

Problem 30: Diketahui himpunan A terdiri atas 4 huruf, yaitu p, q, r, dan s, serta himpunan B terdiri atas 4 angka, yaitu 1, 2, 3, dan 4. Dari kedua himpunan tersebut.

Okay, let's tackle the next problem. We have two sets: Set A with 4 letters (p, q, r, s) and Set B with 4 numbers (1, 2, 3, 4). Now, what exactly are we trying to do with these sets? Without the rest of the question, it's tough to give a complete answer, but I can walk you through some common things we might be asked to do with sets like these.

Possible Questions and Solutions

1. Finding the Number of Possible Functions from A to B

One common question is: How many different functions can we create that map elements from set A to set B? Remember that a function must assign each element in set A to exactly one element in set B.

For each element in A, we have 4 choices in B. Since there are 4 elements in A, the total number of functions is:

4 * 4 * 4 * 4 = 4^4 = 256

So, there are 256 possible functions from A to B.

2. Finding the Number of One-to-One Functions (Injections) from A to B

A one-to-one function, also known as an injection, requires that each element in A maps to a unique element in B. In other words, no two elements in A can map to the same element in B.

  • For the first element in A, we have 4 choices in B.
  • For the second element in A, we only have 3 choices left in B (since we can't reuse the element we used for the first element).
  • For the third element in A, we have 2 choices left in B.
  • For the fourth element in A, we have only 1 choice left in B.

So, the total number of one-to-one functions is:

4 * 3 * 2 * 1 = 4! = 24

There are 24 possible one-to-one functions from A to B.

3. Finding the Number of Onto Functions (Surjections) from A to B

An onto function, also known as a surjection, requires that every element in B is mapped to by at least one element in A. In this case, since A and B have the same number of elements (4), any one-to-one function will also be an onto function, and vice-versa. Therefore, the number of onto functions is the same as the number of one-to-one functions, which is 24.

4. Finding the Number of Bijections from A to B

A bijection is a function that is both one-to-one (injective) and onto (surjective). As we've already established, when the sets have the same number of elements, any injection is also a surjection, and vice versa. So, the number of bijections is the same as the number of one-to-one and onto functions, which is 24.

5. Forming Ordered Pairs

We could also be asked to form ordered pairs from the elements of A and B. For instance, we might want to know how many different ordered pairs can be formed where the first element comes from A and the second element comes from B. In this case, for each of the 4 elements in A, we can pair it with any of the 4 elements in B. So, the total number of ordered pairs is:

4 * 4 = 16

What's Missing?

To give you a precise answer, I need to know what the actual question is asking. Are we looking for the number of functions, one-to-one functions, onto functions, bijections, or something else entirely? Once you provide the full question, I can give you a step-by-step solution. Just let me know!

In the meantime, I hope this explanation of different possibilities helps! Remember, understanding the definitions of these different types of functions (one-to-one, onto, bijection) is key to solving these kinds of problems. Keep practicing, and you'll get the hang of it!

Let me know the full question, and we'll solve it together! Good luck, and happy problem-solving!