Math  /  Algebra

QuestionLet g(x)=3+1.8xg(x)=3+1.8 x and f(x)=3(1.8)xf(x)=3(1.8)^{x}. Explore the rates of change (constant? average?) for the linear function gg and the exponential function ff. a. Determine the value of the ratios Δg(x)Δx\frac{\Delta g(x)}{\Delta x} and Δf(x)Δx\frac{\Delta f(x)}{\Delta x} over the following intervals of xx : a. From x=0x=0 to x=1.5x=1.5. Δg(x)Δx=Δf(x)Δx=\begin{array}{l} \frac{\Delta g(x)}{\Delta x}=\square \\ \frac{\Delta f(x)}{\Delta x}=\square \end{array} b. From x=1.5x=1.5 to x=3x=3. Δg(x)Δx=Δf(x)Δx=\begin{array}{l} \frac{\Delta g(x)}{\Delta x}=\square \\ \frac{\Delta f(x)}{\Delta x}=\square \end{array} c. From x=3x=3 to x=5x=5. Δg(x)Δx=Δf(x)Δx=\begin{array}{l} \frac{\Delta g(x)}{\Delta x}=\square \\ \frac{\Delta f(x)}{\Delta x}=\square \end{array} b. Do the functions gg and ff have constant rates of change? Yes for gg and not for ff because, for any change in xx, the ratio Δg(x)Δx\frac{\Delta g(x)}{\Delta x} is constant and the ratio Δf(x)Δx\frac{\Delta f(x)}{\Delta x} is not constant. No, because, for any change in xx, the ratios Δg(x)Δx\frac{\Delta g(x)}{\Delta x} and Δf(x)Δx\frac{\Delta f(x)}{\Delta x} are not constant. Yes, because, for any change in xx, the ratios Δg(x)Δx\frac{\Delta g(x)}{\Delta x} and Δf(x)Δx\frac{\Delta f(x)}{\Delta x} are constant Yes for ff and no for gg because, for any change in xx, the ratio Δf(x)Δx\frac{\Delta f(x)}{\Delta x} is constant and the

Studdy Solution
a. From x=0x = 0 to x=1.5x = 1.5: Δg(x)Δx=1.8\frac{\Delta g(x)}{\Delta x} = 1.8 and Δf(x)Δx2.82\frac{\Delta f(x)}{\Delta x} \approx 2.82.
From x=1.5x = 1.5 to x=3x = 3: Δg(x)Δx=1.8\frac{\Delta g(x)}{\Delta x} = 1.8 and Δf(x)Δx6.844\frac{\Delta f(x)}{\Delta x} \approx 6.844.
From x=3x = 3 to x=5x = 5: Δg(x)Δx=1.8\frac{\Delta g(x)}{\Delta x} = 1.8 and Δf(x)Δx19.596\frac{\Delta f(x)}{\Delta x} \approx 19.596. b. Yes for gg and not for ff because, for any change in xx, the ratio Δg(x)Δx\frac{\Delta g(x)}{\Delta x} is constant and the ratio Δf(x)Δx\frac{\Delta f(x)}{\Delta x} is not constant.

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