The question asks for the average rate of change of the function f(x) = 100 * 2^x on the interval [0,4]. The average rate of change of a function over the interval [a, b] is given by the formula: \[ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} \] Here, our function f(x) = 100 * 2^x, our interval is [0, 4], so a = 0 and b = 4. We plug these into the function f to get f(0) and f(4): f(0) = 100 * 2^0 = 100 * 1 = 100 f(4) = 100 * 2^4 = 100 * 16 = 1600 Now plug f(0) and f(4) into the rate of change formula: \[ \text{Average Rate of Change} = \frac{f(4) - f(0)}{4 - 0} = \frac{1600 - 100}{4} = \frac{1500}{4} = 375 \] Hence, the average rate of change of f(x) on the interval [0,4] is 375.
The question is asking to compare the average rates of change for the pair of functions over the given interval [1, 5]. The given functions are: - \( f(x) = 9x^2 \) - \( g(x) = 1 + 3x^2 \) The average rate of change of a function over an interval \([a, b]\) can be found using the formula: \[ \text{Average rate of change} = \frac{f(b) - f(a)}{b - a} \] Let's compute this for each function over the interval [1, 5]. For \( f(x) = 9x^2 \): - \( f(1) = 9(1)^2 = 9 \) - \( f(5) = 9(5)^2 = 225 \) - \( \text{Average rate of change of} \ f(x) \text{ over} \ [1, 5] = \frac{f(5) - f(1)}{5 - 1} = \frac{225 - 9}{4} = \frac{216}{4} = 54 \) For \( g(x) = 1 + 3x^2 \): - \( g(1) = 1 + 3(1)^2 = 4 \) - \( g(5) = 1 + 3(5)^2 = 1 + 75 = 76 \) - \( \text{Average rate of change of} \ g(x) \text{ over} \ [1, 5] = \frac{g(5) - g(1)}{5 - 1} = \frac{76 - 4}{4} = \frac{72}{4} = 18 \) Therefore: - The average rate of change of \( f(x) \) over [1, 5] is \( 54 \). - The average rate of change of \( g(x) \) over [1, 5] is \( 18 \). - The average rate of change of \( f(x) - g(x) \) over [1, 5] is \( 54 - 18 = 36 \). So, the average rate of change of \( f \) over [1, 5] is \( 54 \), of \( g \) over [1, 5] is \( 18 \), and of \( f - g \) over [1, 5] is \( 36 \).
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