Using the First Derivative Test, at what point is marginal cost $MC(x)$ at a local minimum?

• A. At a value $x=c$ where $M^C\prime (c)=0$ and $M^C\prime (x)$ changes sign from negative to positive at $c$.
• B. At a value $x=c$ where $M^C\prime (c)\ne 0$ and $M^C\prime (x)$ changes sign from negative to positive at $c$.
• C. At any value $x=c$ where $M^C\prime (c)=0$, regardless of whether the sign of $M^C\prime (x)$ changes.
• D. At a value $x=c$ where $M^C\prime (c)=0$ and $M^C\prime (x)$ changes sign from positive to negative at $c$.

Answer: (A)

Verified step by step guidance

1. Recall that the marginal cost function, denoted as \(M\!C(x)\), represents the derivative of the total cost function with respect to quantity \(x\). To find local minima or maxima of \(M\!C(x)\), we analyze its first derivative, \(M\!C'(x)\).
2. Apply the First Derivative Test: Identify critical points where \(M\!C'(x) = 0\) or where \(M\!C'(x)\) is undefined. These points are candidates for local minima or maxima of the marginal cost function.
3. Examine the sign changes of \(M\!C'(x)\) around each critical point \(x = c\). If \(M\!C'(x)\) changes from negative to positive at \(c\), then \(M\!C(x)\) has a local minimum at \(x = c\).
4. If \(M\!C'(x)\) changes from positive to negative at \(c\), then \(M\!C(x)\) has a local maximum at \(x = c\). If there is no sign change, then \(x = c\) is neither a local minimum nor maximum.
5. Summarize: The marginal cost \(M\!C(x)\) is at a local minimum at \(x = c\) where \(M\!C'(c) = 0\) and \(M\!C'(x)\) changes sign from negative to positive at \(c\).