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