Question

at the end of the holiday season in January, the sales at a department store are expected to fall. It was initially estimated that for the x day of January, The sales will be s(x). The financial analysis at the store correct other projection and are not expecting the total sales for the X day of January to be t(x)T(1)=

Answer 1

`\begin{gathered} T(x)=(13)/(5)+(32)/(5(3x+1)^2) \\ T(1)=\text{ }(13)/(5)+\text{ }(32)/(5(3(1)+1)^2) \\ T(1)=(13)/(5)+(32)/(5(4)^2) \\ T(1)=(13)/(5)+(32)/(5(16)) \\ T(1)=(13)/(5)+(2)/(5) \\ T(1)=(15)/(5)=3 \\ \end{gathered}`
`\begin{gathered} T^(\prime)(x)=\frac{\text{ 13}}{5}+(32)/(5(3x+1)) \\ T^(\prime)(x)=\text{ }(dt(x))/(dx)((13)/(5))+(dt(x))/(dx)((32)/(5(3x+1)) \\ T^(\prime)(x)=0+(\text{ -32\rparen\lparen}((dt(x))/(dx)5(3x+1)^2)/((dt(x))/(dx)(5(3x+1)^2)^2) \\ T^(\prime)(x)=0\text{ - 32\lparen}(5((dt(x))/(dx)(3x+1)^2))/((5(3x+1)^2)^2) \\ T^(\prime)(x)=0\text{ -32\lparen}(5((dt)/(dx)(g)^2)((dt)/(dx)(3x+1))/((5(3x+1)^2)^2) \\ T^(\prime)(x)=0\text{ -32\lparen}(5(2g)(3))/((5(3x+1)^2)^2) \\ T^(\prime)(x)=0\text{ - 32\lparen}^(5(2)(3x+1)(3))/((5(3x+1)^2)^2) \\ T^(\prime)(x)=0\text{ -32\lparen}\frac{5\text{ x \lparen}2)(3x+1)(3)}{25(3x+1)^4} \\ T^(\prime)(x)=0\text{ - 32\lparen}(2(3))/(5(3x+1)^3)) \\ T^(\prime)(x)=0\text{ - 32\lparen}(6)/(5(3x+1)^3)) \\ T^(\prime)(x)=\text{ - }(192)/(5(3x+1)^3) \\ T^(\prime)(1)=\text{ - }(192)/(5(3(1)+1)^3) \\ T^(\prime)(1)=\text{ - }(192)/(5(4)^3) \\ T^(\prime)(1)=\text{ - }(192)/(320) \\ T^(\prime)(1)=\frac{\text{ - 3}}{5} \end{gathered}`