if (f(x)=12 + 20x) and (g(x)=\frac{1}{2}cdot3^{x}), for what values of (x) is (g(x)<f(x))?

Explanation:

Step1: Set up the inequality

Set (g(x)<f(x)), so (\frac{1}{2}\cdot3^{x}<12 + 20x). We can solve this by graphing both functions (y_1=\frac{1}{2}\cdot3^{x}) and (y_2 = 12+20x) and finding the (x -)values where the graph of (y_1) is below the graph of (y_2). Another way is to use a numerical - approximation method. Let's try some values of (x). When (x = 0): (g(0)=\frac{1}{2}\cdot3^{0}=\frac{1}{2}) and (f(0)=12 + 20\times0=12), and (\frac{1}{2}<12). When (x = 1): (g(1)=\frac{1}{2}\cdot3^{1}=\frac{3}{2}) and (f(1)=12 + 20\times1=32), and (\frac{3}{2}<32). When (x = 2): (g(2)=\frac{1}{2}\cdot3^{2}=\frac{9}{2}) and (f(2)=12 + 20\times2=52), and (\frac{9}{2}<52). When (x = 3): (g(3)=\frac{1}{2}\cdot3^{3}=\frac{27}{2}=13.5) and (f(3)=12 + 20\times3=72), and (13.5<72). When (x = 4): (g(4)=\frac{1}{2}\cdot3^{4}=\frac{81}{2}=40.5) and (f(4)=12 + 20\times4=92), and (40.5<92). When (x = 5): (g(5)=\frac{1}{2}\cdot3^{5}=\frac{243}{2}=121.5) and (f(5)=12 + 20\times5=112), and (121.5>112). We can also use a graphing utility or a numerical solver (such as Newton - Raphson method for the function (h(x)=\frac{1}{2}\cdot3^{x}-20x - 12)). By using a graphing calculator, we find that the solution of the inequality (\frac{1}{2}\cdot3^{x}<12 + 20x) is (x<4.77) (approximate value).

Answer:

(x < 4.77) (approximate)