Is \( y=3^{-x} \) an increasing function or a decreasing function? Dncreasing

The function \( y = 3^{-x} \) is a **decreasing** function. Here's why: ### Understanding the Function 1. **Rewriting the Function:** \[ y = 3^{-x} = \left(\frac{1}{3}\right)^x \] This shows that the function is an exponential function with a base between 0 and 1. 2. **Behavior of Exponential Functions:** - **Base > 1:** The function increases as \( x \) increases. - **0 < Base < 1:** The function decreases as \( x \) increases. Since \( \frac{1}{3} \) is between 0 and 1, \( y = \left(\frac{1}{3}\right)^x \) decreases as \( x \) increases. ### Calculating the Derivative To further confirm, let's find the derivative of \( y \) with respect to \( x \): \[ y = 3^{-x} \\ y = e^{-x \ln 3} \\ \frac{dy}{dx} = -\ln 3 \cdot e^{-x \ln 3} = -\ln 3 \cdot 3^{-x} \] Since \( \ln 3 > 0 \) and \( 3^{-x} > 0 \) for all real \( x \), the derivative \( \frac{dy}{dx} \) is negative. A negative derivative indicates that the function is decreasing. ### Graphical Interpretation If you plot \( y = 3^{-x} \), you'll observe that as \( x \) moves from left to right (increasing), the value of \( y \) decreases, reinforcing that the function is decreasing. ### Conclusion \( y = 3^{-x} \) **is a decreasing function** because as the input \( x \) increases, the output \( y \) decreases.