Simplify F(i) For F(x) = X^3 - 2x^2

Hey guys, let's dive into a cool math problem today from the world of functions and complex numbers. We've got a function, $f(x)=x^3-2 x^2$, and we need to figure out which expression is equivalent to $f(i)$. This means we're plugging in the imaginary unit, '$i$', into our function. Remember, '$i$' is defined as the square root of -1, so $i^2 = -1$. This little tidbit is going to be super important as we simplify. The question is asking for an equivalent expression, and we're given four choices: A. $-2+i$, B. $-2-i$, C. $2+i$, and D. $2-i$. To solve this, we'll substitute '$i$' for '$x$' in the given function and then simplify using the properties of complex numbers. It's like a puzzle, and we've got all the pieces right here. We're going to work through it step-by-step, making sure we don't miss any details. So, grab your notebooks, and let's get this done! This problem is a fantastic way to practice working with exponents and imaginary numbers, especially in the context of polynomial functions. Understanding how to evaluate functions with complex inputs is a key skill in many areas of math and science, from electrical engineering to quantum mechanics. So, even though it looks like a simple algebra problem, it's touching on some really fundamental concepts. We'll break down the substitution process, paying close attention to how the powers of '$i$' behave. We know that $i^1 = i$, $i^2 = -1$, $i^3 = i^2 imes i = -1 imes i = -i$, and $i^4 = (i^2)^2 = (-1)^2 = 1$. This pattern repeats every four powers of '$i$', which is a handy trick to remember. For this specific problem, we'll only need up to $i^3$, so we're in good shape. Let's start by writing out the function with '$i$' substituted in. This will be our starting point for all the algebraic manipulation. We'll take our time and ensure every step is clear and logical, so by the end, you'll feel confident in solving similar problems yourself. This isn't just about getting the right answer; it's about understanding the process and building your mathematical toolkit. So, let's get our hands dirty with some serious math!

Alright guys, let's get down to business and substitute '$i$' into our function $f(x)=x^3-2 x^2$. So, we replace every '$x$' with '$i$' to get $f(i) = (i)^3 - 2(i)^2$. Now, the real fun begins with simplifying this expression. We need to evaluate $i^3$ and $i^2$. We already know that $i^2 = -1$. For $i^3$, we can write it as $i^2 imes i$. Since $i^2 = -1$, then $i^3 = -1 imes i = -i$. Now we can substitute these values back into our expression for $f(i)$. So, $f(i) = (-i) - 2(-1)$. Let's simplify this further. $f(i) = -i + 2$. Rearranging this to the standard form of a complex number (real part first, then imaginary part), we get $f(i) = 2 - i$. This looks pretty straightforward, right? We've used the fundamental properties of imaginary numbers and basic algebra to arrive at our answer. The key was remembering the values of $i^2$ and $i^3$. This process of substitution and simplification is a core skill when dealing with polynomial functions and complex numbers. It's like learning the alphabet before you can write sentences. Once you master these basics, you can tackle much more complex mathematical ideas. Think about it: we took an abstract function and evaluated it at a specific, non-real number. This ability to work with numbers beyond the real number line opens up a whole new world of mathematical possibilities. The problem essentially tests your understanding of the definition of '$i$' and how it interacts with algebraic operations like exponentiation and multiplication. By breaking down $i^3$ into $i^2 imes i$, we were able to utilize the known value of $i^2$ to find the value of $i^3$. This is a common strategy when simplifying higher powers of '$i$'. You'll often see patterns emerge, and recognizing these patterns is crucial for efficiency. In this case, the pattern $i, -1, -i, 1$ is key. Since we only needed $i^2$ and $i^3$, we were able to solve it quickly. The final step of rearranging to the standard $a+bi$ form ensures our answer is presented clearly and matches the format of the multiple-choice options. So, we have $f(i) = 2 - i$. Let's quickly check our options to see if this matches any of them. Yes, it does! Option D is $2-i$. So, we've successfully found the equivalent expression for $f(i)$. This problem is a great example of how seemingly complex mathematical expressions can be simplified by understanding and applying fundamental definitions and rules. It’s all about taking it one step at a time and not getting intimidated by the symbols. Keep practicing, and you’ll be a pro in no time!

So, to recap, we were asked to find the expression equivalent to $f(i)$ where $f(x) = x^3 - 2x^2$. We began by substituting '$i$' for '$x$' in the function, which gave us $f(i) = i^3 - 2i^2$. The critical step here was recalling or deriving the values of the powers of '$i$'. We know that $i^2 = -1$. Then, $i^3$ can be expressed as $i^2 imes i$, which equals $(-1) imes i$, or simply $-i$. With these values in hand, we plugged them back into our expression: $f(i) = (-i) - 2(-1)$. Simplifying this equation, we first handle the multiplication: $-2(-1)$ becomes $+2$. So, the expression transforms into $f(i) = -i + 2$. Finally, to present this in the standard complex number form ($a+bi$, where '$a$' is the real part and '$b$' is the imaginary part), we rearrange the terms to get $f(i) = 2 - i$. This result directly matches one of the given options. Specifically, it's option D. This problem really underscores the importance of understanding the basic properties of imaginary numbers. Without knowing that $i^2 = -1$ and subsequently how to evaluate $i^3$, we wouldn't be able to simplify the expression. The process itself is a testament to the elegance of algebra: manipulate symbols according to defined rules and arrive at a simplified, equivalent form. It's like following a recipe; each step has a purpose, and if you follow it correctly, you get the desired outcome. For anyone just getting into complex numbers, this type of problem is an excellent starting point. It's concrete and allows you to see the abstract concept of '$i$' in action. Many students find imaginary numbers a bit strange at first because they don't correspond to anything we can easily visualize on a number line like real numbers do. However, their utility is immense, and mastering them opens doors to advanced mathematics and applied sciences. Think of this as building a foundation. A strong foundation in evaluating functions with complex inputs will serve you well as you encounter more advanced topics like complex analysis, Fourier transforms, and signal processing. So, don't shy away from these problems, guys! Embrace them as opportunities to deepen your mathematical understanding. The fact that we could substitute '$i$' into a polynomial, which is a very common mathematical structure, and get a clear, simple answer like $2-i$ shows how consistent and powerful the rules of algebra are, even when we extend them beyond real numbers. It's a beautiful thing, really. We've gone from a function definition to a specific complex number value through a series of logical steps. This methodical approach is what math is all about.

Let's quickly summarize the final answer and confirm its validity. We started with the function $f(x) = x^3 - 2x^2$. Our goal was to find the expression equivalent to $f(i)$, where '$i$' is the imaginary unit. By substituting '$i$' for '$x$', we obtained $f(i) = (i)^3 - 2(i)^2$. The key to solving this lies in understanding the powers of '$i$'. We know that $i^2 = -1$. Consequently, $i^3 = i^2 imes i = -1 imes i = -i$. Substituting these values back into the expression for $f(i)$, we get $f(i) = (-i) - 2(-1)$. Simplifying this expression involves performing the multiplication first: $-2 imes -1 = +2$. Thus, the expression becomes $f(i) = -i + 2$. Finally, we write this in the standard form of a complex number, $a+bi$, which is $2 - i$. This result, $2-i$, directly corresponds to option D. Therefore, the expression equivalent to $f(i)$ is $2-i$. This problem serves as a great reminder of the foundational rules governing imaginary numbers and their application within polynomial functions. It highlights that the standard algebraic operations apply consistently, even when dealing with complex numbers. For those new to this topic, remember that the cyclical nature of powers of '$i$' ($i^1=i, i^2=-1, i^3=-i, i^4=1$, and then it repeats) is a fundamental concept that simplifies many calculations. Mastering these basic powers is essential for efficiently evaluating expressions involving complex numbers. This exercise is not just about finding a numerical answer; it's about building confidence and proficiency in working with the complex number system, which is an indispensable tool in numerous scientific and engineering disciplines. By systematically applying the rules of algebra, we successfully navigated the substitution and simplification process. The journey from the function definition to the specific complex number $2-i$ is a clear demonstration of algebraic principles in action. It's this methodical approach that allows mathematicians and scientists to solve complex problems and make new discoveries. So, keep practicing, keep exploring, and you'll find that the world of mathematics, including complex numbers, is full of fascinating insights and powerful applications.

To wrap things up, guys, we've successfully evaluated $f(i)$ for the function $f(x) = x^3 - 2x^2$. The process involved a direct substitution of '$i$' into the function, leading to $f(i) = i^3 - 2i^2$. The crucial step was utilizing the known values of the powers of '$i$': $i^2 = -1$ and $i^3 = -i$. Plugging these into our expression gave us $f(i) = -i - 2(-1)$. Simplifying this further, we got $f(i) = -i + 2$. Rewriting this in the standard $a+bi$ format, we arrived at our final answer: $f(i) = 2 - i$. This matches option D. This problem is a fantastic little workout for your understanding of imaginary numbers. It shows you how to take a function and plug in a complex number, then simplify the result. It's super important to remember those powers of '$i$' ($i, -1, -i, 1$) because they pop up all over the place in math and science. Whether you're dealing with electrical circuits, quantum mechanics, or just solving some tough algebra problems, knowing how to handle '$i$' is key. The beauty of math is that even when we step outside the familiar realm of real numbers, the rules of algebra still hold true. This consistency allows us to build complex theories and models. So, next time you see a problem like this, don't sweat it! Just substitute, simplify using the powers of '$i$', and rearrange into the standard form. You've got this! Keep practicing, and you'll be a master of complex numbers in no time. It's all about building that confidence and solidifying your understanding, one problem at a time.

Final Answer: The final answer is oxed{2-i}