Mastering Even Functions: What If `g(x)` Is Odd?
What Exactly Are Even and Odd Functions, Anyway?Alright, let's kick things off by making sure we're all on the same page about what even and odd functions actually are. Trust me, guys, these definitions are the bedrock of our entire discussion, so getting them down pat is super important. Think of them as special kinds of symmetry that functions can have. A function f(x) is called an even function if, for every x in its domain, f(-x) = f(x). What does that fancy math talk mean in plain English? It means if you plug in a negative version of a number, you get the exact same output as if you plugged in the positive version of that number. Graphically, this creates a beautiful symmetry around the y-axis. Imagine folding your graph paper along the y-axis; if the two halves match up perfectly, you've got yourself an even function! Classic examples include f(x) = x^2, f(x) = x^4, or f(x) = cos(x). Take f(x) = x^2 for instance: f(2) = 2^2 = 4 and f(-2) = (-2)^2 = 4. See? Same output! This property makes even functions incredibly useful in many areas, like physics when dealing with symmetrical forces or in signal processing for specific types of signals. They're predictable, consistent, and generally quite well-behaved. Understanding this fundamental definition of an even function is our first big win!
Now, let's flip the coin and talk about odd functions. A function g(x) is defined as an odd function if, for every x in its domain, g(-x) = -g(x). This one's a little different. Here, when you plug in a negative number, the output you get is the negative of the output you'd get if you plugged in the positive number. Graphically, odd functions exhibit symmetry about the origin. This means if you rotate the graph 180 degrees around the point (0,0), it will look exactly the same. Think of it this way: if you reflect it across the y-axis and then reflect it across the x-axis, you get the original graph back. Cool, right? Some prime examples of odd functions are g(x) = x^3, g(x) = x^5, or g(x) = sin(x). Let's use g(x) = x^3: g(2) = 2^3 = 8, and g(-2) = (-2)^3 = -8. Notice how g(-2) is -g(2). This relationship, g(-x) = -g(x), is absolutely key for our problem today because our starting point is an odd function g(x). These functions are equally important, appearing in scenarios like wave functions or the behavior of certain electrical signals. So, to recap: even means f(-x) = f(x) (y-axis symmetry), and odd means g(-x) = -g(x) (origin symmetry). Keep these definitions in your mental toolkit, because we're about to put them to work!
Diving Deep: The Power of g(x) as an Odd FunctionOkay, now that we're pros at defining even and odd functions, let's really focus on the core of our problem: we're given that g(x) is an odd function. This isn't just a minor detail; it's the superpower we'll be leveraging to solve our puzzle! Remember, the definition of an odd function, g(-x) = -g(x), is our guiding star here. This means that no matter what g(x) looks like (as long as it's odd), we can always substitute g(-x) with -g(x) whenever we see it. This simple, yet incredibly powerful, algebraic manipulation is what will allow us to test each candidate function and determine its parity (whether it's even, odd, or neither). It's like having a secret decoder ring for functions!
So, why is this so important? Because we're trying to find a function f(x) that must be even. To prove a function f(x) is even, we need to show that f(-x) = f(x). Our strategy will be to take each given option, substitute -x into it, and then use our odd function superpower (g(-x) = -g(x)) to simplify the expression. After simplification, we'll compare the result with the original f(x). If they match, boom! We've found our even function. If they don't, then that option is out. It's a systematic and logical approach that guarantees we'll uncover the correct answer. This process highlights the beauty of formal definitions in math—they provide a clear pathway to proving or disproving properties. Understanding that g(x) being odd means every instance of g(-x) transforms into -g(x) is the key. Without this fundamental understanding, we'd be lost in a sea of variables. So, let's keep that g(-x) = -g(x) firmly in mind as we move on to scrutinize each option presented to us. Get ready to put on your detective hats; we're about to uncover the truth!
Option A: f(x) = g(x) + 2 - Is It Even?Alright, let's kick off our investigation with the first contender: f(x) = g(x) + 2. Remember, our mission is to determine if this function is even. To do that, we need to check if f(-x) is equal to f(x). If it is, awesome! If not, then this one's a bust. So, let's plug -x into our f(x) and see what happens.
Here's the breakdown:
- Start with
f(-x): We replace everyxinf(x)with-x. So,f(-x) = g(-x) + 2. - Apply the odd function property: This is where our knowledge about
g(x)being an odd function comes into play. We know thatg(-x)is equivalent to-g(x). So, we can substitute that right in:f(-x) = -g(x) + 2. - Compare
f(-x)withf(x): Now, let's look at what we have forf(-x):-g(x) + 2. And what's our originalf(x)? It'sg(x) + 2. Are these two expressions the same? Nope, not at all! Unlessg(x)itself happens to be zero (which isn't always the case for any odd function),-g(x) + 2is definitely not equal tog(x) + 2. For example, ifg(x) = x(which is an odd function), thenf(x) = x + 2. Following our calculation,f(-x) = -x + 2. Clearly,x + 2is not the same as-x + 2for most values ofx. If they were equal, it would meanx + 2 = -x + 2, which simplifies tox = -x, implying2x = 0, orx = 0. This only holds true for a single point, not for the entire function. Therefore, this function,f(x) = g(x) + 2, is not generally an even function. It fails our even function test becausef(-x)does not equalf(x). This combination often results in a function that is neither even nor odd, unlessg(x)is identically zero, which is a trivial case. So, option A is out!
Option B: f(x) = g(x) + g(x) - Double Trouble or Even Goodness?Next up on our function parade is Option B: f(x) = g(x) + g(x). Before we dive into the even function test, let's make this expression a little simpler, shall we? g(x) + g(x) is just 2g(x). So, our function is effectively f(x) = 2g(x). Now, with this simplified form, let's get down to business and see if f(-x) matches f(x). Remember the drill: plug in -x, use the odd function property of g(x), and then compare!
Here’s how it unfolds:
- Start with
f(-x): We substitute-xintof(x) = 2g(x). This gives usf(-x) = 2g(-x). - Apply the odd function property: Since we know
g(x)is an odd function, we can confidently replaceg(-x)with-g(x). So, our expression becomesf(-x) = 2(-g(x)). - Simplify
f(-x): Multiplying that out, we getf(-x) = -2g(x). - Compare
f(-x)withf(x): Now for the big reveal! We havef(-x) = -2g(x). And what was our originalf(x)? It was2g(x). Are2g(x)and-2g(x)the same? Only ifg(x)is zero, which, again, isn't always the case. In fact,f(-x) = -f(x)in this scenario. This, my friends, is the definition of an odd function! So, whilef(x) = 2g(x)is a perfectly valid function, it doesn't meet our criteria for being an even function. It's actually another odd function! For example, ifg(x) = x^3, thenf(x) = 2x^3. Let's check:f(2) = 2(2^3) = 16, andf(-2) = 2((-2)^3) = 2(-8) = -16. Clearly,f(-2) = -f(2). So,f(x) = 2g(x)inherits the oddness fromg(x). This option is also a no-go for our quest to find an even function. Option B is also out of the running!
Option C: f(x) = g(x)^2 - The Winner Revealed!Alright, folks, it's time for the moment of truth! We've evaluated two options, and they didn't quite make the cut for being an even function. Now, let's turn our attention to Option C: f(x) = g(x)^2. This one looks a little different, as it involves squaring our odd function g(x). Could this be the one? Let's follow our rigorous process to find out. Our goal, as always, is to see if f(-x) equals f(x).
Here’s the breakdown, step-by-step:
- Start with
f(-x): We replace everyxinf(x) = g(x)^2with-x. This gives usf(-x) = (g(-x))^2. Notice how theg(-x)part is inside the parentheses, and then the entire result ofg(-x)is squared. This distinction is super important for the next step. - Apply the odd function property: This is where the magic happens! Since
g(x)is an odd function, we know its definition:g(-x) = -g(x). So, we can substitute-g(x)in place ofg(-x)within our expression forf(-x). This transforms our equation intof(-x) = (-g(x))^2. - Simplify
f(-x): Now, let's simplify(-g(x))^2. When you square a negative term, what happens? That's right, the negative sign disappears! Squaring any term, whether it's positive or negative, always results in a positive or non-negative value. So,(-g(x))^2simply becomesg(x)^2. Therefore,f(-x) = g(x)^2. - Compare
f(-x)withf(x): Now for the big reveal! We found thatf(-x) = g(x)^2. And what was our original functionf(x)? It was alsog(x)^2. Bingo! We have a match! Sincef(-x) = f(x), this function definitely fits the definition of an even function.
This is a super cool property, guys! When you take an odd function and square it, you're essentially canceling out its oddness because the act of squaring removes any negative signs introduced by the odd function property. Think of an example: let g(x) = x, which is an odd function (g(-x) = -x). If we apply our option C, f(x) = g(x)^2 = (x)^2 = x^2. And what do we know about f(x) = x^2? It's the classic parabola, perfectly symmetrical about the y-axis, making it an undeniable even function! Another example: g(x) = sin(x), an odd function. Then f(x) = (sin(x))^2 = sin^2(x). We know sin(-x) = -sin(x), so sin^2(-x) = (-sin(x))^2 = sin^2(x). See? It works! So, after careful consideration and a bit of algebraic wizardry, we can confidently declare that f(x) = g(x)^2 is indeed the even function we were looking for. Option C is our winner!
Option D: f(x) = -g(x) - Another Odd One Out?Alright, we're down to our last contender, Option D: f(x) = -g(x). We've already found our winner in Option C, but for completeness and to solidify our understanding, let's put this one through the paces as well. Remember, our goal is to see if f(-x) is equal to f(x). If it is, then it's an even function. If not, then it's something else.
Let's break it down:
- Start with
f(-x): We replace everyxinf(x) = -g(x)with-x. This gives usf(-x) = -g(-x). - Apply the odd function property: Here comes our trusty odd function definition for
g(x)! We know thatg(-x)is equivalent to-g(x). So, we can substitute that into our expression. This transformsf(-x) = -g(-x)intof(-x) = -(-g(x)). - Simplify
f(-x): What happens when you have a double negative? They cancel each other out, becoming a positive! So,-(-g(x))simply simplifies tog(x). Therefore,f(-x) = g(x). - Compare
f(-x)withf(x): Now, let's compare! We've found thatf(-x) = g(x). And what was our original functionf(x)? It was-g(x). Areg(x)and-g(x)the same? No way! Unlessg(x)is identically zero, these two expressions are negatives of each other. In fact, what we've found is thatf(-x) = -f(x)(sinceg(x)is the negative of-g(x)). This, my friends, is the definition of an odd function!
So, just like Option B, this function f(x) = -g(x) turns out to be an odd function, not an even one. Multiplying an odd function by a negative constant simply