Mathematics, 10.10.2019 04:30, puffley
Find an equation of the tangent line to the graph of the function f(x)=(x^3+1)(3x^2-4x+2) at the point (1,2)
Answers: 3
![\bf f(x)=(x^3+1)(3x^2-4x+2) \\\\\\ \cfrac{df}{dx}=3x^2(3x^2-4x+2)+(x^3+1)(6x-4)\impliedby \textit{product rule} \\\\\\ \left. \cfrac{df}{dx} \right|_{x=1}\implies 3(1)^2[3(1)^2-4(1)+2]~+~(1^3+1)[6(1)-4] \\\\\\ 3[3-4+2]~+~(2\cdot 2)\implies \stackrel{slope}{7}\\\\ -------------------------------\\\\ \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-2=7(x-1)\implies y-2=7x-7 \\\\\\ y=7x-5](/tpl/images/0305/8568/05798.png)
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Find an equation of the tangent line to the graph of the function f(x)=(x^3+1)(3x^2-4x+2) at the poi...
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