1. The problem asks for the least value of $n$ such that the $n$-th derivative of the function $f(x) = x^4 - 2x^3 + 3x^2 - 4x + 5$ is zero. 2. Recall that the derivative of a polynomial reduces the degree by 1 each time. The $n$-th derivative of a polynomial of degree $m$ is zero if $n > m$. 3. Since $f(x)$ is a polynomial of degree 4, the first derivative $f'(x)$ is degree 3, the second derivative $f''(x)$ is degree 2, the third derivative $f^{(3)}(x)$ is degree 1, the fourth derivative $f^{(4)}(x)$ is degree 0 (a constant), and the fifth derivative $f^{(5)}(x)$ is zero. 4. Let's compute the derivatives step-by-step: $$f(x) = x^4 - 2x^3 + 3x^2 - 4x + 5$$ $$f'(x) = 4x^3 - 6x^2 + 6x - 4$$ $$f''(x) = 12x^2 - 12x + 6$$ $$f^{(3)}(x) = 24x - 12$$ $$f^{(4)}(x) = 24$$ $$f^{(5)}(x) = 0$$ 5. The least $n$ for which $f^{(n)}(x) = 0$ is $n=5$. Final answer: (c) 5