Select the correct answer. Exponential function f is represented by the table. x & -2 & -1 & 0 & 1 & 2 f(x) & -62 & -30 & -14 & -6 & -2 Function g is represented by the equation. [ g(x)=-20((1)/(2))^x+10 ] Which statement correctly compares the two functions on the interval [-2,1] ? A. Both functions are increasing, but function g increases at a faster average ra B. Only function f is increasing, and both functions are negative. C. Both functions are increasing, but function f increases at a faster average rate D. Only function f is increasing, and only function f is negative.

## Step 1: Analyze Function \( f \)<br />### The function \( f(x) \) is represented by the table. We observe the values of \( f(x) \) for \( x = -2, -1, 0, 1, 2 \). The values are \( -62, -30, -14, -6, -2 \), respectively. As \( x \) increases from \(-2\) to \(1\), \( f(x) \) also increases, indicating that \( f(x) \) is an increasing function on the interval \([-2, 1]\).<br /><br />## Step 2: Analyze Function \( g \)<br />### The function \( g(x) = -20\left(\frac{1}{2}\right)^{x} + 10 \) is given. We need to evaluate this function at \( x = -2, -1, 0, 1 \):<br />- For \( x = -2 \): <br /> \[<br /> g(-2) = -20\left(\frac{1}{2}\right)^{-2} + 10 = -20 \times 4 + 10 = -80 + 10 = -70<br /> \]<br />- For \( x = -1 \): <br /> \[<br /> g(-1) = -20\left(\frac{1}{2}\right)^{-1} + 10 = -20 \times 2 + 10 = -40 + 10 = -30<br /> \]<br />- For \( x = 0 \): <br /> \[<br /> g(0) = -20\left(\frac{1}{2}\right)^{0} + 10 = -20 \times 1 + 10 = -20 + 10 = -10<br /> \]<br />- For \( x = 1 \): <br /> \[<br /> g(1) = -20\left(\frac{1}{2}\right)^{1} + 10 = -20 \times \frac{1}{2} + 10 = -10 + 10 = 0<br /> \]<br /><br />### From these calculations, we see that \( g(x) \) is also increasing on the interval \([-2, 1]\).<br /><br />## Step 3: Compare the Functions<br />### Both functions \( f \) and \( g \) are increasing on the interval \([-2, 1]\). However, we need to compare their average rates of increase. The average rate of increase for a function over an interval can be calculated as the change in function value divided by the change in \( x \).<br /><br />- For \( f(x) \) from \( x = -2 \) to \( x = 1 \):<br /> \[<br /> \text{Average rate of increase for } f = \frac{f(1) - f(-2)}{1 - (-2)} = \frac{-6 - (-62)}{3} = \frac{56}{3} \approx 18.67<br /> \]<br /><br />- For \( g(x) \) from \( x = -2 \) to \( x = 1 \):<br /> \[<br /> \text{Average rate of increase for } g = \frac{g(1) - g(-2)}{1 - (-2)} = \frac{0 - (-70)}{3} = \frac{70}{3} \approx 23.33<br /> \]<br /><br />### Comparing the average rates, \( g(x) \) increases at a faster average rate than \( f(x) \).