The marginal cost in dollars of producing x units is given by the next equation:
![C=2.6\sqrt[]{x}+600](https://img.qammunity.org/2023/formulas/mathematics/college/xmz7qrdqum6qq9er6uwg9f51o7wxak3cwj.png)
a)
To find the marginal cost (in dollars per unit) when x= 9.
Then, we need to replace x=9 on the derivation of the cost equation:
So:
![(d)/(dx)C=\frac{1.3}{\sqrt[]{x}}](https://img.qammunity.org/2023/formulas/mathematics/college/yren0c7is1n1soampxd29lpghzhb02hu6b.png)
Where:
![(d)/(dx)2.6\sqrt[]{x}=2.6(d)/(dx)\sqrt[]{x}=2.6(d)/(dx)^{}x^{(1)/(2)}=2.6\cdot(1)/(2)x^{(1)/(2)-1}=1.3\cdot x^{-(1)/(2)}=\frac{1.3}{\sqrt[]{3}}](https://img.qammunity.org/2023/formulas/mathematics/college/8udvic3anyiee5ske2ue667fh7r7xmhb4h.png)
and, the derivate of a constant is equal to zero.

Replacing x= 9
![(d)/(dx)C=\frac{1.3}{\sqrt[]{9}}](https://img.qammunity.org/2023/formulas/mathematics/college/5e1or9fjc7ade1xrz9y3zj48vbmddfdlxt.png)
Hence, the marginal cost is equal to:

b) Now, when the production increases 9 to 10. It's the same as the cost of producing one more machine beyond 9.
Then, it would be x=10 on the cost equation:
![C=2.6\sqrt[]{x}+600](https://img.qammunity.org/2023/formulas/mathematics/college/xmz7qrdqum6qq9er6uwg9f51o7wxak3cwj.png)
![C=2.6\sqrt[]{10}+600](https://img.qammunity.org/2023/formulas/mathematics/college/8gb8e0mar5yxozgsp869ansdrccnlctx4h.png)

and x= 9
![C=2.6\sqrt[]{9}+600](https://img.qammunity.org/2023/formulas/mathematics/college/cqoeqrzn4ry3sl4ubukywoj27hv73jsgz2.png)


Then, we calculate C(10) - C(9) =


C)
Both results are equal.
Hence, the marginal cost when x=9 is equal to the additional cost when the production increases from 9 to 10.