Comparing Function Values & Finding Domain: F(x) = X^(-40)
Hey guys! Today, we're diving into the world of functions, specifically looking at the function f(x) = x^(-40). We're going to tackle two main tasks: comparing the values of this function at different points and figuring out its domain. So, buckle up, and let's get started!
Comparing Function Values for f(x) = x^(-40)
Our first challenge is to compare the function values at various points. Remember, f(x) = x^(-40) can also be written as f(x) = 1 / x^(40). This form is super helpful because it clearly shows that as the absolute value of x increases, the value of f(x) decreases, and vice versa, as long as x is not zero. This is because we're dealing with an even power (40), which means the result will always be positive regardless of whether x is positive or negative.
1) Comparing f(18) and f(16)
Let's kick things off by comparing f(18) and f(16). We need to figure out which one is larger. Since f(x) = 1 / x^(40), we're essentially comparing 1 / 18^(40) and 1 / 16^(40). The key here is that as the denominator (x^(40)) gets bigger, the overall fraction gets smaller. Because 18 is greater than 16, 18^(40) is much, much greater than 16^(40). Therefore, 1 / 18^(40) is smaller than 1 / 16^(40). So, we can confidently say that f(16) > f(18). This might seem counterintuitive at first, but remember the inverse relationship due to the negative exponent!
2) Comparing f(-42) and f(2.5)
Next up, we have f(-42) and f(2.5). Now, this one is interesting because we're comparing a value with a large negative input to a small positive input. Since we have an even power, (-42)^(40) will be a huge positive number, and f(-42) = 1 / (-42)^(40) will be a very tiny positive fraction (almost zero). On the other hand, f(2.5) = 1 / (2.5)^(40) will also be a positive fraction, but since 2.5 is much smaller in absolute value than -42, (2.5)^(40) will be a much smaller positive number than (-42)^(40). This means 1 / (2.5)^(40) will be significantly larger than 1 / (-42)^(40). Hence, f(2.5) > f(-42).
3) Comparing f(-1.6) and f(-1.7)
Now, let's compare f(-1.6) and f(-1.7). Both inputs are negative, and we need to think about their absolute values. The absolute value of -1.7 is greater than the absolute value of -1.6. Therefore, (-1.7)^(40) will be greater than (-1.6)^(40). Consequently, 1 / (-1.7)^(40) will be smaller than 1 / (-1.6)^(40). Thus, f(-1.6) > f(-1.7).
4) Comparing f(-8) and f(6)
Moving on, we're comparing f(-8) and f(6). Again, we have a negative and a positive input. Since the exponent is even, the negative sign won't matter when we raise -8 to the power of 40. So, we're essentially comparing 1 / (-8)^(40) and 1 / 6^(40), which is the same as comparing 1 / 8^(40) and 1 / 6^(40). Because 8 is greater than 6, 8^(40) is greater than 6^(40), and thus 1 / 8^(40) is smaller than 1 / 6^(40). Therefore, f(6) > f(-8).
5) Comparing f(24) and f(-24)
Finally, we have f(24) and f(-24). This is a cool one! Since we have an even exponent, raising 24 or -24 to the power of 40 will yield the same result. That is, 24^(40) = (-24)^(40). Therefore, 1 / 24^(40) is equal to 1 / (-24)^(40). So, f(24) = f(-24). This highlights the symmetry of the function due to the even exponent.
Finding the Domain of the Function f(x) = x^(-40)
Okay, now let's switch gears and figure out the domain of our function, f(x) = x^(-40). The domain of a function is simply the set of all possible input values (x-values) for which the function is defined. In other words, it's all the x values we can plug into the function without causing any mathematical mayhem!
We know that f(x) = x^(-40) can be rewritten as f(x) = 1 / x^(40). This form gives us a crucial clue about the domain. We can see that we're dividing by x^(40). Now, remember the golden rule of division: we can't divide by zero! So, the only value that x cannot be is 0. If x were 0, we'd have 1 / 0^(40), which is undefined.
For all other real numbers, x^(40) is perfectly well-defined, and so is 1 / x^(40). Therefore, the domain of f(x) = x^(-40) is all real numbers except for 0. We can express this in a few ways:
- Set notation: { x | x ∈ ℝ, x ≠ 0 }
- Interval notation: (-∞, 0) ∪ (0, ∞)
Both of these notations tell us the same thing: x can be any number from negative infinity up to 0 (but not including 0), and then from 0 (not including 0) to positive infinity.
Wrapping Up
So, there you have it! We've successfully compared the values of the function f(x) = x^(-40) at various points and determined its domain. Remember, understanding the behavior of functions, like how they change with different inputs and what their domains are, is a fundamental skill in algebra and calculus. Keep practicing, and you'll become a function whiz in no time! Keep an eye on the negative exponents, and the inverse relationship they create and always remember you cannot divide by 0, which will help you master domains. You've got this!