suppose that the cost, in dollars, for a company to produce $x$ pairs of a new line of jeans is $c(x)=6800 +…

Explanation:

Step1: Differentiate the cost function

The power - rule for differentiation is $\frac{d}{dx}(x^n)=nx^{n - 1}$, and the derivative of a constant $c$ is 0. $C(x)=6800 + 5x+0.01x^{2}+0.0002x^{3}$ $C'(x)=\frac{d}{dx}(6800)+\frac{d}{dx}(5x)+\frac{d}{dx}(0.01x^{2})+\frac{d}{dx}(0.0002x^{3})$ $C'(x)=0 + 5+0.01\times2x+0.0002\times3x^{2}=5 + 0.02x+0.0006x^{2}$

Step2: Evaluate $C'(x)$ at $x = 100$

Substitute $x = 100$ into $C'(x)$: $C'(100)=5+0.02\times100 + 0.0006\times100^{2}$ $C'(100)=5 + 2+0.0006\times10000$ $C'(100)=5 + 2+6=13.00$

Step3: Evaluate $C(x)$ at $x = 100$

Substitute $x = 100$ into $C(x)$: $C(100)=6800+5\times100+0.01\times100^{2}+0.0002\times100^{3}$ $C(100)=6800 + 500+0.01\times10000+0.0002\times1000000$ $C(100)=6800+500 + 100+200$ $C(100)=7600.00$

Answer:

a. $C'(x)=5 + 0.02x+0.0006x^{2}$ b. $C'(100)=13.00$ dollars/pair c. $C(100)=7600.00$ dollars