The given equation is
[tex]C=\frac{1014000}{Q}+5.8Q[/tex]
To find the increase from 353 to 354, substitute Q by these values, then subtract the answers
[tex]\begin{gathered} Q=353 \\ I.C=\lbrack\frac{1014000}{354}+5.8(354)\rbrack-\lbrack\frac{1014000}{353}+5.8(353)\rbrack \end{gathered}[/tex]
Calculate it
[tex]\begin{gathered} I\mathrm{}C=\lbrack4917.60678\rbrack-\lbrack4919.921246\rbrack \\ I\mathrm{}C=-2.314466459 \end{gathered}[/tex]
Round it to 2 decimal places
I.C = -2.31
Now we will find C' using derivative
[tex]\begin{gathered} C=1014000Q^{-1}+5.8Q \\ C^{\prime}=1014000(-1)Q^{-1-1}+5.8(1)Q^{1-1} \end{gathered}[/tex]
Simplify each term
[tex]\begin{gathered} C^{\prime}=-1014000Q^{-2}+5.8Q^0 \\ C^{\prime}=-\frac{1014000}{Q^2}+5.8 \end{gathered}[/tex]
Substitute Q by 353
[tex]\begin{gathered} C^,=-\frac{1014000}{(353)^2}+5.8 \\ C^{\prime}=-2.337453956 \end{gathered}[/tex]
Round it to the nearest 2 decimal place
[tex]C^{\prime}(353)=-2.34[/tex]
The answer is
Change in C = -2.31
C'(353) = -2.34
