Q.The total cost function of a firm is $C(x) = x^3 - 3x^2 + 5x + 100$, where $x$ is the number of units produced. Find the marginal cost and the average cost when $x=4$.

Marginal cost is the derivative of $C(x)$ evaluated at $x=4$; average cost is $C(4)$ divided by $4$.

$MC(4) = 3(16)-6(4)+5 = 48-24+5=29$. $C(4)=64-48+20+100=136$, so $AC(4)=136/4=34$.

Step 1 — Differentiate to find marginal cost. $C(x) = x^3-3x^2+5x+100 \Rightarrow MC = C'(x) = 3x^2-6x+5$.

Step 2 — Evaluate MC at $x=4$. $MC(4) = 3(4)^2-6(4)+5 = 3(16)-24+5 = 48-24+5=29$.

Step 3 — Find total cost at $x=4$. $C(4) = (4)^3-3(4)^2+5(4)+100 = 64-48+20+100=136$.

Step 4 — Find average cost at $x=4$. $AC(4) = \dfrac{C(4)}{4} = \dfrac{136}{4}=34$.

Independent check. Re-computing $C(4)$ term-by-term separately: $4^3=64$; $-3(4)^2=-3(16)=-48$; $5(4)=20$; constant $100$. Sum: $64-48+20+100=136$ ✓, matching Step 3. Also, since $MC(4)=29$ is noticeably less than $AC(4)=34$, average cost must currently be falling as output rises past $x=4$ (average cost falls whenever $MC<AC$) — consistent with a firm still on the downward-sloping part of its average cost curve.

Confusing marginal cost (the derivative $C'(x)$) with average cost ($C(x)/x$) is the single most common error — remember MC is the cost of the next unit, AC is the cost per unit averaged over all units made so far, and the two are generally different numbers except at AC's own minimum point.