Mastering Functions: Finding Values And Domains

Hey guys! Today, we're diving deep into the awesome world of functions in mathematics. We'll be tackling two super important skills: finding the value of a function for specific inputs (that's part (a)!) and figuring out the domain of a function (part (b)!). Don't sweat it if this sounds a bit intimidating; we'll break it all down nice and easy. Think of a function like a cool machine: you put something in (an input, or 'x'), and it gives you something out (an output, or 'f(x)'). Our job is to understand what goes in and what comes out.

We'll be working through a few examples, like $f(x) = x^3$, $f(x) = 2x - 1$, $f(x) = x^2 - 3x$, and $f(x) = 5 + x - 2x^2$. For each, we'll plug in specific numbers for 'x' and see what 'f(x)' we get. We'll also explore the 'domain' of each function. The domain is basically all the possible 'x' values that you can legally plug into the function without breaking it. It's like knowing all the ingredients your function machine can handle!

So, grab your calculators, maybe a comfy seat, and let's get ready to become function wizards. Whether you're in a high school math class, prepping for college, or just love a good math challenge, this guide is for you. We're aiming to make these concepts crystal clear, so you can confidently tackle any function problem that comes your way. Let's get started!

Understanding Functions: The Input-Output Machine

Alright, let's kick things off by really getting a handle on what a function is. In mathematics, a function is a rule that assigns to each input value exactly one output value. You can think of it like a vending machine. You press a button (the input), and a specific item (the output) comes out. There's no guesswork; one button always gives you the same snack. In math terms, we use 'x' to represent our input and 'f(x)' to represent the output. The notation $f(x)$ is read as "f of x," and it signifies the value of the function 'f' when the input is 'x'.

The domain of a function is crucial because it tells us which 'x' values are allowed. For some functions, you can plug in any real number, and that's awesome! For others, there might be restrictions. For example, you can't divide by zero, and you can't take the square root of a negative number (in the realm of real numbers, anyway). Understanding these limitations is key to correctly defining the domain. We often express the domain using inequalities (like $x eq 3$) or interval notation (like $(- eq, 3) ext{ U } (3, eq)$).

Let's look at our first example: $f(x) = x^3$. This function is pretty straightforward. It means "take the input 'x' and multiply it by itself three times." The beauty of $f(x) = x^3$ is that you can cube any real number – positive, negative, or zero – and get a valid result. There are no divisions by zero, no square roots of negatives. This means the domain for $f(x) = x^3$ is all real numbers. We can write this as $- eq < x < eq$ or, in interval notation, $(- eq, eq)$.

Now, consider $f(x) = 2x - 1$. This function tells us to take our input 'x', multiply it by 2, and then subtract 1. Again, you can perform these operations on any real number. There are no inherent restrictions. So, the domain for $f(x) = 2x - 1$ is also all real numbers. This is a common characteristic of linear functions – they are defined for all real inputs.

Things get a little more interesting with functions like $f(x) = x^2 - 3x$. Here, we square the input and then subtract three times the input. Are there any numbers you can't square? Nope. Are there any numbers you can't multiply by 3? Nope. So, again, this polynomial function has a domain of all real numbers. The same applies to $f(x) = 5 + x - 2x^2$. Since this is also a polynomial (just a quadratic one), it can accept any real number as an input without issues. The domain is all real numbers.

So, the big takeaway here is that for many common functions, especially polynomials, the domain is often all real numbers. But it's always good practice to think about potential restrictions, especially when you see fractions or square roots. We'll explore those more in future discussions, but for now, let's focus on applying these concepts to our specific problems!

Problem 49: Cubing It Up with $f(x)=x^3$

Alright, let's dive into our first problem, which features the function $f(x) = x^3$. This is a classic cubic function, meaning the highest power of 'x' is 3. It's like taking your input number and multiplying it by itself twice more. For example, if your input is 2, you do $2 imes 2 imes 2$, which equals 8. If your input is -3, you do $(-3) imes (-3) imes (-3)$, which equals -27. See? It works for both positive and negative numbers.

(a) Find $f(x)$ for the indicated values of $x$.

We are asked to find the value of $f(x)$ when $x = -2$ and when $x = 5$. Let's tackle these one by one.

• When $x = -2$: To find $f(-2)$, we substitute -2 for every 'x' in the function $f(x) = x^3$. So, $f(-2) = (-2)^3$. Calculating this, we get $(-2) imes (-2) imes (-2)$. $(-2) imes (-2) = 4$ $4 imes (-2) = -8$ Therefore, $f(-2) = -8$. Pretty neat, huh?

• When $x = 5$: Now, let's find $f(5)$. We substitute 5 for 'x' in $f(x) = x^3$. So, $f(5) = (5)^3$. Calculating this, we get $5 imes 5 imes 5$. $5 imes 5 = 25$ $25 imes 5 = 125$ Therefore, $f(5) = 125$. Easy peasy!

(b) Find the domain of $f$. Use words or inequalities to describe the domain.

The domain of a function is the set of all possible input values ('x' values) for which the function is defined. For $f(x) = x^3$, can we think of any real number that we cannot cube? Nope! You can cube positive numbers, negative numbers, and zero. The result will always be a real number. There are no divisions by zero or square roots of negative numbers to worry about here.

So, the domain of $f(x) = x^3$ is all real numbers. We can express this in a few ways:

• In words: The domain is all real numbers.
• Using inequalities: $- eq < x < eq$
• Using interval notation: $(- eq, eq)$

All these notations mean the exact same thing: you can plug in any real number you want for 'x' into this function.

Problem 50: Linear Power with $f(x)=2x-1$

Next up, we have a linear function, $f(x) = 2x - 1$. Linear functions are super common and represent straight lines when graphed. This particular function tells us to take our input 'x', multiply it by 2, and then subtract 1 from the result.

(a) Find $f(x)$ for the indicated values of $x$.

We need to find the output of the function when $x = 8$ and when $x = -1$.

• When $x = 8$: Let's substitute 8 for 'x' in our function $f(x) = 2x - 1$. $f(8) = 2(8) - 1$ First, we multiply: $2 imes 8 = 16$. Then, we subtract: $16 - 1 = 15$. So, $f(8) = 15$. That's our answer for this input!

• When $x = -1$: Now, let's plug in -1 for 'x'. Remember to be careful with the negative signs! $f(-1) = 2(-1) - 1$ First, multiply: $2 imes (-1) = -2$. Then, subtract: $-2 - 1 = -3$. So, $f(-1) = -3$. Watch out for those negatives, guys!

(b) Find the domain of $f$. Use words or inequalities to describe the domain.

The domain for $f(x) = 2x - 1$ is the set of all possible 'x' values. Think about the operations involved: multiplication by 2 and subtraction by 1. Can you perform these operations on any real number? Yes, you absolutely can! There are no denominators that could be zero, and no square roots of negative numbers. This means that this linear function is defined for every single real number.

Therefore, the domain of $f(x) = 2x - 1$ is all real numbers. We can write this as:

• In words: The domain is all real numbers.
• Using inequalities: $- eq < x < eq$
• Using interval notation: $(- eq, eq)$

Linear functions are generally very accommodating when it comes to their domains!

Problem 51: Quadratic Expressions with $f(x)=x^2-3x$

Let's move on to a quadratic function, $f(x) = x^2 - 3x$. Quadratic functions have a highest power of 'x' as 2, and they typically graph as parabolas. This function involves squaring the input and then subtracting three times the input.

(a) Find $f(x)$ for the indicated values of $x$.

We need to calculate the function's value for $x = -3$ and $x = 2$.

• When $x = -3$: Substitute -3 for 'x' in $f(x) = x^2 - 3x$. $f(-3) = (-3)^2 - 3(-3)$ Remember that squaring a negative number makes it positive: $(-3)^2 = (-3) imes (-3) = 9$. And multiplying negatives gives a positive: $-3(-3) = 9$. So, the expression becomes $9 - (-9)$, which is $9 + 9 = 18$. Therefore, $f(-3) = 18$. Nice work!

• When $x = 2$: Now, let's plug in $x = 2$. $f(2) = (2)^2 - 3(2)$ First, square 2: $(2)^2 = 4$. Then, multiply: $3(2) = 6$. So, the expression is $4 - 6$. $4 - 6 = -2$. Therefore, $f(2) = -2$. You're getting the hang of this!

(b) Find the domain of $f$. Use words or inequalities to describe the domain.

For the function $f(x) = x^2 - 3x$, we need to determine what 'x' values are allowed. This function involves squaring 'x' and multiplying 'x' by 3, then subtracting. Can you square any real number? Yes. Can you multiply any real number by 3? Yes. Can you subtract these results? Yes. There are no operations here that would cause problems for any real input.

This means the domain of $f(x) = x^2 - 3x$ is all real numbers. Polynomial functions, like this quadratic, are defined for every real number.

• In words: The domain is all real numbers.
• Using inequalities: $- eq < x < eq$
• Using interval notation: $(- eq, eq)$

Problem 52: Another Quadratic Adventure with $f(x)=5+x-2x^2$

Our final function is another quadratic function: $f(x) = 5 + x - 2x^2$. This one has a constant term (the 5), a linear term (the x), and a squared term (the $-2x^2$). It's just a slightly different arrangement than the previous quadratic, but the principles are the same.

(a) Find $f(x)$ for the indicated values of $x$.

We are not given specific values of 'x' to evaluate for this problem in the prompt. However, if we were given values, the process would be identical to the previous examples. For instance, let's pick a couple of values to demonstrate, say $x=1$ and $x=-4$. Remember, the prompt asks to find $f(x)$ for indicated values, and since none were indicated for #52, we can't complete this part as written. However, for illustrative purposes:

• If $x = 1$: $f(1) = 5 + (1) - 2(1)^2$ $f(1) = 5 + 1 - 2(1)$ $f(1) = 5 + 1 - 2$ $f(1) = 6 - 2$ $f(1) = 4$

• If $x = -4$: $f(-4) = 5 + (-4) - 2(-4)^2$ $f(-4) = 5 - 4 - 2(16)$ $f(-4) = 1 - 32$ $f(-4) = -31$

(b) Find the domain of $f$. Use words or inequalities to describe the domain.

Now, let's talk about the domain of $f(x) = 5 + x - 2x^2$. This function is a polynomial function (specifically, a quadratic). Polynomials are defined for all real numbers. You can add, subtract, and multiply any real numbers, and you can raise any real number to a non-negative integer power (like squaring it). None of the operations in $f(x) = 5 + x - 2x^2$ will ever result in an undefined value for any real number 'x'.

Therefore, the domain of $f(x) = 5 + x - 2x^2$ is all real numbers. We can state this as:

• In words: The domain is all real numbers.
• Using inequalities: $- eq < x < eq$
• Using interval notation: $(- eq, eq)$

Every polynomial function, regardless of its degree, shares this property of having a domain that encompasses all real numbers.

Wrapping It Up: Function Fluency!

So there you have it, guys! We've successfully navigated through finding function values for specific inputs and determining the domains of various functions. We saw that for $f(x) = x^3$, $f(-2) = -8$ and $f(5) = 125$. For $f(x) = 2x - 1$, we found $f(8) = 15$ and $f(-1) = -3$. And for $f(x) = x^2 - 3x$, we calculated $f(-3) = 18$ and $f(2) = -2$. For $f(x) = 5 + x - 2x^2$, we illustrated how to find values if they were indicated.

Crucially, we also established that for all these functions ($f(x) = x^3$, $f(x) = 2x - 1$, $f(x) = x^2 - 3x$, and $f(x) = 5 + x - 2x^2$), the domain is all real numbers. This is because these are all polynomial functions, and they don't have any inherent restrictions like division by zero or taking the square root of a negative number.

Remember, the domain is all about what you can plug in. Practice makes perfect, so try working through more examples on your own. Pay close attention to any functions that involve fractions (where the denominator can't be zero) or square roots (where the expression inside can't be negative). Those are the scenarios where the domain might be restricted. Keep practicing, and you'll be a function pro in no time! Stay sharp!