Given the piecewise function defined below, answer the questions at the bottom of the page regarding the continuity and differentiability of $f$ at $x=1$.
\[
f(x)=\left\{\begin{array}{lll}
2 e^{3 x-3}+1 & \text { for } & x \leq 1 \\
6 x^{2}-3 x & \text { for } & x> 1
\end{array}\right.
\]

Step 1 ：To make the function continuous, both expressions must have the same value when \(x=1\).

Step 2 ：For the first expression, when \(x=1\), we have \(f(1)=2e^{3(1)-3}+1=2e^0+1=2(1)+1=3\).

Step 3 ：For the second expression, when \(x=1\), we have \(f(1)=6(1)^2-3(1)=6-3=3\).

Step 4 ：Since both expressions have the same value when \(x=1\), the function is continuous at \(x=1\).

Step 5 ：Now, we need to check the differentiability of the function at \(x=1\).

Step 6 ：For the first expression, the derivative is \(f'(x)=6e^{3x-3}\) for \(x\le 1\).

Step 7 ：For the second expression, the derivative is \(f'(x)=12x-3\) for \(x>1\).

Step 8 ：Now, we need to check if the derivatives have the same value when \(x=1\).

Step 9 ：For the first expression, when \(x=1\), we have \(f'(1)=6e^{3(1)-3}=6e^0=6(1)=6\).

Step 10 ：For the second expression, when \(x=1\), we have \(f'(1)=12(1)-3=12-3=9\).

Step 11 ：Since the derivatives do not have the same value when \(x=1\), the function is not differentiable at \(x=1\).

Step 12 ：\(\boxed{\text{The function is continuous but not differentiable at } x=1}\)