If $f^{\prime}(x)=15 x^{2}-4$ and $f(x)$ passes through the point $(-2,3)$ then $f(-1)=$

Step 1 ：Given that the derivative of the function, \(f'(x) = 15x^2 - 4\), we can find the original function by integrating \(f'(x)\).

Step 2 ：The integral of \(f'(x)\) is \(f(x) = 5x^3 - 4x + c\), where \(c\) is the constant of integration.

Step 3 ：We know that the function passes through the point \((-2,3)\). Substituting these values into the function gives us the equation \(3 = 5(-2)^3 - 4(-2) + c\).

Step 4 ：Solving this equation for \(c\) gives us \(c = 35\).

Step 5 ：Substituting \(c = 35\) back into the function gives us \(f(x) = 5x^3 - 4x + 35\).

Step 6 ：We are asked to find the value of the function at \(x = -1\). Substituting \(x = -1\) into the function gives us \(f(-1) = 5(-1)^3 - 4(-1) + 35\).

Step 7 ：Solving this equation gives us \(f(-1) = 34\).

Step 8 ：So, the final answer is \(f(-1) = \boxed{34}\).