Question

Use the definition of “\(f(x)\) is \(O(g(x))\)” to show that \({2^x} + 17\) is \(O({3^x})\).

Short answer

By the definition of “\(f(x)\)is \(O(g(x))\)”, \({2^x} + 17\)is\(O({3^x})\)with \(C = 2\) and \(k = 2\).

Step-by-step solution

Step 1:

According to the definition,

\(f(x)\)is\(O(g(x))\)if there are positive constants\(C\)and\(k\)such that,

\(\left| {f(x)} \right| \ge C\left| {g(x)} \right|\),

whenever \(x > k\).

Step 2:

Taking\(x > 1\)we get the property\({3^x} > {2^x}\).

We know that,

\(\begin{array}{l}{3^2} = 9\\{3^3} = 27\end{array}\)

Thus we get the property,

\({3^x} > 17\) when \(x > 2\),

 Step 3:

To show, we now take\(k = 2\)and we get\(x > 2\),

\(\begin{array}{l}\left| {f(x)} \right| = \left| {{2^x} + 17} \right|\\\therefore 0 \le {2^x} + 17\\ \le {3^x} + 17\\ \le {3^x} + {3^x}\end{array}\)

\(\begin{array}{l} = {2.3^x}\\ = 2\left| {{3^x}} \right|\end{array}\)

Therefore we get \(C = 2\) and \(k = 2\).