Mastering Function Operations: A Step-by-Step Guide

h1: Mastering Function Operations: A Step-by-Step Guide

Welcome to our guide on mastering function operations! Today, we're diving deep into how to combine and evaluate functions, a fundamental skill in mathematics that unlocks a world of problem-solving possibilities. You'll learn how to tackle problems involving the addition, subtraction, and scaling of functions. Let's start with a clear understanding of the functions we'll be working with. We have two distinct functions: $h(x) = x^2 + 1$ and $k(x) = x - 2$. Understanding these individual functions is the first step before we can perform operations on them.

• $h(x) = x^2 + 1$: This is a quadratic function. When you input any value for 'x', you first square it, and then you add 1 to the result. For example, if $x=2$, $h(2) = 2^2 + 1 = 4 + 1 = 5$. If $x=3$, $h(3) = 3^2 + 1 = 9 + 1 = 10$. The 'x' here represents any input value, and $h(x)$ is the output value of the function.
• $k(x) = x - 2$: This is a linear function. For any input 'x', you simply subtract 2 from it. For instance, if $x=2$, $k(2) = 2 - 2 = 0$. If $x=3$, $k(3) = 3 - 2 = 1$. Just like $h(x)$, 'x' is the input, and $k(x)$ is the output.

These functions are the building blocks for our exploration. By understanding what each function does individually, we can confidently move on to combining them and evaluating the results. This process is not just about crunching numbers; it's about understanding the relationships between different mathematical expressions and how they behave when put together. It's like learning the individual notes before you can play a melody. We'll be using these specific functions throughout our examples, so keep them in mind as we proceed. The concepts we cover, however, are broadly applicable to any pair of functions you encounter.

Understanding $(h+k)(x)$ and $(h-k)(x)$

Now, let's explore how to combine these functions. The notation $(h+k)(x)$ means we're creating a new function by adding the outputs of $h(x)$ and $k(x)$ together for any given input 'x'. Similarly, $(h-k)(x)$ means we're creating another new function by subtracting the output of $k(x)$ from the output of $h(x)$ for any given input 'x'. It's crucial to remember that we are operating on the function definitions first, and then we evaluate the resulting combined function at a specific point. This is different from evaluating each function separately and then combining the results, although in some cases, it can lead to the same answer.

• $(h+k)(x)$: To find the definition of $(h+k)(x)$, we simply add the expressions for $h(x)$ and $k(x)$.

  $$(h+k)(x) = h(x) + k(x)$$

  Substituting our given functions:

  $$(h+k)(x) = (x^2 + 1) + (x - 2)$$

  Now, we simplify by combining like terms:

  $$(h+k)(x) = x^2 + x + 1 - 2$$

  $$(h+k)(x) = x^2 + x - 1$$

  This new expression, $x^2 + x - 1$, is the definition of the combined function $(h+k)(x)$.

• $(h-k)(x)$: To find the definition of $(h-k)(x)$, we subtract the expression for $k(x)$ from $h(x)$. It's important to use parentheses around $k(x)$ to ensure we subtract the entire expression, especially when it contains multiple terms.

  $$(h-k)(x) = h(x) - k(x)$$

  Substituting our given functions:

  $$(h-k)(x) = (x^2 + 1) - (x - 2)$$

  Distribute the negative sign to each term inside the parentheses:

  $$(h-k)(x) = x^2 + 1 - x + 2$$

  Now, we simplify by combining like terms:

  $$(h-k)(x) = x^2 - x + 1 + 2$$

  $$(h-k)(x) = x^2 - x + 3$$

  This new expression, $x^2 - x + 3$, is the definition of the combined function $(h-k)(x)$.

Understanding these combined function definitions is key to solving the specific evaluation problems. We've successfully created two new functions from our original ones. The next step is to see how these new functions behave when we plug in specific numbers.

Evaluating $(h+k)(2)$

Now that we have the combined function definition for $(h+k)(x)$, we can evaluate it at a specific point. We need to find the value of $(h+k)(2)$. This means we take our combined function $(h+k)(x) = x^2 + x - 1$ and substitute $x=2$ into it.

$$(h+k)(2) = (2)^2 + (2) - 1$$

First, calculate the exponent:

$$(h+k)(2) = 4 + 2 - 1$$

Next, perform the addition:

$$(h+k)(2) = 6 - 1$$

Finally, perform the subtraction:

$$(h+k)(2) = 5$$

So, the value of $(h+k)(2)$ is 5. This means that when you input 2 into the function $h(x)$ and 2 into the function $k(x)$, and then add their results, you get 5. Let's quickly verify this by evaluating $h(2)$ and $k(2)$ separately and adding them:

• $h(2) = (2)^2 + 1 = 4 + 1 = 5$
• $k(2) = 2 - 2 = 0$
• $h(2) + k(2) = 5 + 0 = 5$

As you can see, the result matches! This confirms our understanding of how $(h+k)(x)$ works. It's a powerful way to simplify calculations when dealing with operations on functions.

Evaluating $(h-k)(3)$

Similarly, we can evaluate $(h-k)(3)$ using the combined function definition we derived: $(h-k)(x) = x^2 - x + 3$. We need to substitute $x=3$ into this expression.

$$(h-k)(3) = (3)^2 - (3) + 3$$

First, calculate the exponent:

$$(h-k)(3) = 9 - 3 + 3$$

Next, perform the subtraction:

$$(h-k)(3) = 6 + 3$$

Finally, perform the addition:

$$(h-k)(3) = 9$$

Thus, the value of $(h-k)(3)$ is 9. This signifies that if you take the value of $h(3)$ and subtract the value of $k(3)$, the result is 9. Let's verify this by calculating $h(3)$ and $k(3)$ separately:

• $h(3) = (3)^2 + 1 = 9 + 1 = 10$
• $k(3) = 3 - 2 = 1$
• $h(3) - k(3) = 10 - 1 = 9$

Once again, the results match, reinforcing our grasp of function subtraction. This method allows for a streamlined approach to complex calculations involving function manipulation.

Evaluating $3h(2) + 2k(3)$

Our final evaluation involves a combination of scalar multiplication and addition of function values at different points. We need to calculate $3h(2) + 2k(3)$. This means we will evaluate $h(x)$ at $x=2$, multiply that result by 3, then evaluate $k(x)$ at $x=3$, multiply that result by 2, and finally, add these two scaled results together.

First, let's find $h(2)$. We already did this earlier, but let's recall:

$h(2) = (2)^2 + 1 = 4 + 1 = 5$

Now, multiply $h(2)$ by 3:

$3h(2) = 3 imes 5 = 15$

Next, let's find $k(3)$. We also calculated this before:

$k(3) = 3 - 2 = 1$

Now, multiply $k(3)$ by 2:

$2k(3) = 2 imes 1 = 2$

Finally, add the two scaled results:

$3h(2) + 2k(3) = 15 + 2 = 17$

So, the value of $3h(2) + 2k(3)$ is 17. This type of problem highlights how we can manipulate function outputs in various ways, including scaling them by constants and summing them up. It shows the versatility of functions in mathematical modeling and problem-solving.

Conclusion

We've successfully navigated through various operations involving functions $h(x)$ and $k(x)$. We learned how to define new functions $(h+k)(x)$ and $(h-k)(x)$ by adding and subtracting the original functions, respectively. Then, we applied these definitions to evaluate specific points, finding that $(h+k)(2) = 5$ and $(h-k)(3) = 9$. Finally, we tackled a more complex expression, $3h(2) + 2k(3)$, which resulted in 17. These operations are fundamental in calculus, algebra, and many other areas of mathematics. Mastering them will provide a solid foundation for more advanced topics. Keep practicing these skills, and don't hesitate to explore different function pairs and operations.

For further exploration into the fascinating world of functions and their applications, you can visit Khan Academy's comprehensive resources on algebra and functions. Their interactive lessons and practice problems are an excellent way to deepen your understanding. Another valuable resource for mathematical concepts is Wolfram MathWorld, which offers detailed explanations and definitions for a vast array of mathematical topics.