To approximate \(3^{\sqrt{3}}\) using the function \(f(x) = 3^x\), we can create a table with \(x\) values that converge on \(\sqrt{3}\) (approximately \(1.732\)). We'll evaluate \(f(x)\) for a series of \(x\) values that approach \(\sqrt{3}\) from both below and above.
Table of \(f(x) = 3^x\)
| \(x\) | \(f(x) = 3^x\) | |---------------|-------------------| | 1.7 | \(3^{1.7} \approx 4.989\) | | 1.71 | \(3^{1.71} \approx 5.024\) | | 1.72 | \(3^{1.72} \approx 5.053\) | | 1.73 | \(3^{1.73} \approx 5.083\) | | 1.74 | \(3^{1.74} \approx 5.113\) | | 1.75 | \(3^{1.75} \approx 5.144\) | | 1.76 | \(3^{1.76} \approx 5.174\) | | 1.77 | \(3^{1.77} \approx 5.205\) | | 1.78 | \(3^{1.78} \approx 5.235\) | | 1.79 | \(3^{1.79} \approx 5.266\) | | 1.8 | \(3^{1.8} \approx 5.297\) | | 1.73 (more accurate) | \(3^{1.732} \approx 5.081\) |
Conclusion
From the values calculated, we can observe the behavior as we approach \(x = \sqrt{3} \approx 1.732\). The value of \(3^{\sqrt{3}}\) is approximately \(5.081\) when \(x\) is at \(1.732\). Hence, to three decimal places:
\[ 3^{\sqrt{3}} \approx 5.081 \]