The function \(f(x)\) is defined as:
\[ f(x) = \begin{cases} 3-x, & \text{if } x \leq 0 \\ x^2 + 2, & \text{if } x \geq 0 \end{cases} \]
Since \(-3 \leq 0\), we use the first piece of the function: \(f(x) = 3-x\).
Substitute \(x = -3\) into the expression:
\[f(-3) = 3 - (-3) = 3 + 3 = 6\]
Thus, the value of \(f(-3)\) is 6.
The product of all solutions of the equation \(e^{5(\log_e x)^2 + 3 = x^8, x > 0}\) , is :