Domain And Range Of Exponential Function F(x)=(1/5)^x

by Alex Johnson 54 views

When we delve into the world of functions, two fundamental concepts that we always explore are the domain and the range. Understanding these helps us grasp the behavior and limitations of a function. Today, we're going to unravel the mysteries of the domain and range for a specific exponential function: f(x)=(15)xf(x) = \left(\frac{1}{5}\right)^x. This function might look a little different from the typical exponential functions you might have encountered, like 2x2^x or exe^x, because its base is a fraction less than one. But don't let that fool you; the principles remain the same, and exploring its domain and range will solidify your understanding of these crucial mathematical ideas. Let's break down what these terms mean in general before we apply them to our particular function. The domain of a function refers to the set of all possible input values (the 'x' values) for which the function is defined and produces a valid output. Think of it as the set of all 'allowed' numbers you can plug into the function. On the other hand, the range of a function is the set of all possible output values (the 'f(x)' or 'y' values) that the function can produce. It's the collection of all results you can get after plugging in valid inputs. For many functions, like linear functions, the domain and range are often all real numbers. However, for other types of functions, such as square root functions or rational functions, there can be restrictions. Exponential functions, in particular, have unique characteristics that define their domain and range. Let's now focus on our specific function, f(x)=(15)xf(x) = \left(\frac{1}{5}\right)^x, and see how these concepts apply.

Let's dive deeper into determining the domain of f(x)=(15)xf(x) = \left(\frac{1}{5}\right)^x. The domain is essentially the set of all valid 'x' values we can substitute into the function without causing any mathematical issues, like division by zero or taking the square root of a negative number. For exponential functions, especially those in the form axa^x where 'a' is a positive constant not equal to 1, there are generally no restrictions on the exponent 'x'. This means you can raise a positive base to any real number power – positive, negative, or zero – and you will always get a real number as a result. Consider our base, 15\frac{1}{5}. This is a positive number and it's not equal to 1. Therefore, we can plug in any real number for 'x' and the expression (15)x\left(\frac{1}{5}\right)^x will be perfectly well-defined. For instance, if x=2x=2, f(2)=(15)2=125f(2) = \left(\frac{1}{5}\right)^2 = \frac{1}{25}. If x=−1x=-1, f(−1)=(15)−1=5f(-1) = \left(\frac{1}{5}\right)^{-1} = 5. If x=0x=0, f(0)=(15)0=1f(0) = \left(\frac{1}{5}\right)^0 = 1. If x=0.5x=0.5, f(0.5)=(15)0.5=15f(0.5) = \left(\frac{1}{5}\right)^{0.5} = \frac{1}{\sqrt{5}}. You can see that no matter what real number we choose for 'x', we always get a defined real number output. There are no values of 'x' that would lead to undefined operations. Therefore, the domain of f(x)=(15)xf(x) = \left(\frac{1}{5}\right)^x is the set of all real numbers. We can express this mathematically as x∈Rx \in \mathbb{R}, or in interval notation as (−∞,∞)(-\infty, \infty). This understanding is crucial because it tells us that the function is defined for every single point along the horizontal axis.

Now, let's shift our focus to the range of f(x)=(15)xf(x) = \left(\frac{1}{5}\right)^x. The range represents the set of all possible output values, or 'y' values, that the function can produce. To figure this out, let's consider the behavior of our exponential function. The base of our function is 15\frac{1}{5}, which is a positive number less than 1. This means that as the value of 'x' increases, the value of (15)x\left(\frac{1}{5}\right)^x will decrease, approaching zero. For example, if x=1x=1, f(1)=15f(1) = \frac{1}{5}. If x=2x=2, f(2)=125f(2) = \frac{1}{25}. If x=10x=10, f(10)=(15)10f(10) = \left(\frac{1}{5}\right)^{10}, which is a very, very small positive number. As 'x' approaches positive infinity (x→∞x \to \infty), f(x)f(x) approaches 0 (f(x)→0f(x) \to 0). However, it never actually reaches 0. This is because you can't raise a positive number to any finite power and get exactly zero. On the other hand, let's consider what happens when 'x' becomes very small (approaches negative infinity, x→−∞x \to -\infty). In this case, the value of \left(\frac{1}{5} ight)^x will become very large. For instance, if x=−1x=-1, f(−1)=5f(-1) = 5. If x=−2x=-2, f(−2)=25f(-2) = 25. If x=−10x=-10, f(−10)=(15)−10=510f(-10) = \left(\frac{1}{5}\right)^{-10} = 5^{10}, which is a huge positive number. So, as 'x' approaches negative infinity, f(x)f(x) approaches positive infinity (f(x)→∞f(x) \to \infty). Crucially, throughout all these values of 'x', the output f(x)f(x) is always positive. It never becomes zero or negative. Therefore, the range of f(x)=(15)xf(x) = \left(\frac{1}{5}\right)^x is all real numbers strictly greater than zero. We can express this mathematically as f(x)>0f(x) > 0, or in interval notation as (0,∞)(0, \infty). This tells us that the graph of the function will always lie above the x-axis.

To summarize our findings for f(x)=(15)xf(x) = \left(\frac{1}{5}\right)^x: the domain is the set of all real numbers, meaning we can input any real number for 'x'. This is represented as x∈Rx \in \mathbb{R} or (−∞,∞)(-\infty, \infty). The range is the set of all real numbers greater than zero, meaning the output of the function will always be a positive value. This is represented as f(x)>0f(x) > 0 or (0,∞)(0, \infty). This behavior is characteristic of exponential functions with a base between 0 and 1. The graph of this function would start very high on the left side (as 'x' approaches negative infinity) and decrease as 'x' increases, getting closer and closer to the x-axis but never touching or crossing it, eventually approaching zero as 'x' approaches positive infinity. This specific characteristic, where the function approaches a certain value (in this case, 0) but never reaches it, is known as a horizontal asymptote. For f(x)=(15)xf(x) = \left(\frac{1}{5}\right)^x, the horizontal asymptote is the line y=0y=0 (the x-axis). Understanding the domain and range is vital for analyzing functions, graphing them accurately, and solving related mathematical problems. It gives us a complete picture of the function's behavior and its possible values.

For further exploration into the fascinating world of functions and their properties, you might find it helpful to visit resources that offer in-depth explanations and examples. A great place to start is with Khan Academy, which provides comprehensive lessons and practice exercises on domains, ranges, and various types of functions. Another excellent resource is the Wolfram MathWorld website, which offers detailed mathematical definitions and explanations, including those related to exponential functions and their properties. These sites can provide you with additional perspectives and tools to deepen your understanding.