Question

Show that \({x^2} + 4x + 17\) is \(O({x^3})\)but that \({x^3}\) is not \(O({x^2} + 4x + 17)\).

Short answer

Hence, we obtain \({x^2} + 4x + 17\) is \(O({x^3})\)but \({x^3}\) is not \(O({x^2} + 4x + 17)\)

Step-by-step solution

Step 1:

Let\(f(x) = {x^3}\)is\(O({x^2} + 4x + 17)\)

Then,\(\left| {{x^3}} \right| \le C\left| {{x^2} + 4x + 17} \right|\)where\(C\)and\(k\)is constant

When we take\(x > k\)

Assuming\(k > 17\)

\( \Rightarrow x > 17\)

Step 2:

Using property,\(\left| {a + b} \right| \le \left| a \right| + \left| b \right|\)

Here, we are getting a contradiction because\(\left| {{x^3}} \right| > \left| {3C{x^2}} \right|\)when\(x > 3C\)

\( \Rightarrow \)\(f(x) = {x^3}\) is not \(O({x^2} + 4x + 17)\)