Domain & Range Of Exponential Function K(x) = -2^x

by Andrew McMorgan 51 views

Hey guys, let's dive into the exciting world of exponential functions and figure out the domain and range of a specific function, k(x)=βˆ’2xk(x) = -2^x. We know that kk is a transformation of the parent exponential function, f(x)=2xf(x) = 2^x. Understanding these transformations is key to unlocking the secrets of functions, so buckle up!

Understanding the Parent Function: f(x) = 2^x

Before we tackle k(x)=βˆ’2xk(x) = -2^x, it's super important to get a solid grip on its parent function, f(x)=2xf(x) = 2^x. This is your basic, go-to exponential function. When we talk about the domain of f(x)=2xf(x) = 2^x, we're asking ourselves: "What values of xx can we plug into this function without breaking it?" For exponential functions like this, there are no restrictions on the xx-values. You can plug in any real number – positive, negative, zero, fractions, decimals, you name it! So, the domain of f(x)=2xf(x) = 2^x is all real numbers, which we can write as (βˆ’βˆž,∞)(-\infty, \infty) or x∈R{x \in R}. Now, let's think about the range. The range is all about the possible output values, the yy-values, that the function can produce. For f(x)=2xf(x) = 2^x, since the base (2) is positive, the output will always be positive. No matter what real number you raise 2 to, the result will never be zero or negative. It will always be greater than zero. So, the range of f(x)=2xf(x) = 2^x is (0,∞)(0, \infty) or y∈R∣y>0{y \in R \mid y > 0}. Keep this in mind, as it's the foundation for understanding our transformed function.

Transforming the Parent Function: k(x) = -2^x

Alright, let's get down to business with our function k(x)=βˆ’2xk(x) = -2^x. This function is a transformation of the parent function f(x)=2xf(x) = 2^x. Specifically, the negative sign in front of the 2x2^x indicates a reflection across the x-axis. Think about what this reflection does to the graph of y=2xy = 2^x. Every positive yy-value from the parent function becomes a negative yy-value in the transformed function. For instance, if f(3)=23=8f(3) = 2^3 = 8, then k(3)=βˆ’23=βˆ’8k(3) = -2^3 = -8. This transformation is crucial for determining the domain and range of k(x)k(x).

Digging into the Domain of k(x) = -2^x

Now, let's nail down the domain of k(x)=βˆ’2xk(x) = -2^x. Remember, the domain is about the possible input values for xx. Did our transformation (the reflection across the x-axis) add any restrictions to the values of xx we can plug in? Nope! Just like with the parent function f(x)=2xf(x) = 2^x, we can still plug in any real number for xx into k(x)=βˆ’2xk(x) = -2^x. The negative sign only affects the output, not the input. So, the domain of k(x)k(x) remains all real numbers. We can express this as (βˆ’βˆž,∞)(-\infty, \infty) or x∈R{x \in R}. It's a common misconception that the negative sign might limit the domain, but in this case, it doesn't.

Unpacking the Range of k(x) = -2^x

The range is where the real magic of this transformation happens. The range refers to the set of all possible output yy-values. We know that the parent function f(x)=2xf(x) = 2^x has a range of y∈R∣y>0{y \in R \mid y > 0}. That means all its outputs are positive. When we apply the transformation k(x)=βˆ’2xk(x) = -2^x, which is a reflection across the x-axis, we are essentially taking all those positive yy-values and making them negative. So, if f(x)f(x) produced values like 1, 2, 4, 8, etc., then k(x)k(x) will produce values like -1, -2, -4, -8, etc. This means that the output of k(x)k(x) will always be negative. It will never be zero, and it will certainly never be positive. Therefore, the range of k(x)=βˆ’2xk(x) = -2^x is all negative real numbers. We can write this as (βˆ’βˆž,0)(-\infty, 0) or y∈R∣y<0{y \in R \mid y < 0}. This is a direct consequence of the reflection across the x-axis.

Comparing with the Options

Let's look at the options provided:

A. Domain: x∈Rβˆ£βˆ’βˆž<x<∞{x \in R \mid -\infty < x < \infty}, Range: y∈R∣y≀0{y \in R \mid y \leq 0} B. Domain: x∈Rβˆ£βˆ’βˆž<x<∞{x \in R \mid -\infty < x < \infty}, Range: y∈R∣y<0{y \in R \mid y < 0}

Based on our analysis, the domain of k(x)=βˆ’2xk(x) = -2^x is indeed all real numbers, which is correctly represented as x∈Rβˆ£βˆ’βˆž<x<∞{x \in R \mid -\infty < x < \infty}.

Now for the range. We determined that the range consists of all negative real numbers, meaning yy must be strictly less than 0. Option A states the range as y∈R∣y≀0{y \in R \mid y \leq 0}, which includes 0. However, as we established, the function k(x)=βˆ’2xk(x) = -2^x will never output 0. Option B states the range as y∈R∣y<0{y \in R \mid y < 0}, which perfectly matches our findings. The outputs are all negative real numbers, excluding 0.

Conclusion: The Correct Answer

So, the correct domain and range for the function k(x)=βˆ’2xk(x) = -2^x are:

  • Domain: x∈Rβˆ£βˆ’βˆž<x<∞{x \in R \mid -\infty < x < \infty} (all real numbers)
  • Range: y∈R∣y<0{y \in R \mid y < 0} (all negative real numbers)

This means Option B is the correct answer, guys! Keep practicing these transformations, and you'll be a function master in no time. It's all about understanding how changes to the basic function affect its input and output possibilities. The reflection across the x-axis is a powerful transformation that flips the graph vertically, directly impacting the range by inverting all the output values. Remember this concept for future problems!