Question

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)

Answer 1

$\bf f(x)=(x^3+1)(3x^2-4x+2) \\\\\\ \cfrac{df}{dx}=3x^2(3x^2-4x+2)+(x^3+1)(6x-4)\impliedby \textit{product rule} \\\\\\ \left. \cfrac{df}{dx} \right|_{x=1}\implies 3(1)^2[3(1)^2-4(1)+2]~+~(1^3+1)[6(1)-4] \\\\\\ 3[3-4+2]~+~(2\cdot 2)\implies \stackrel{slope}{7}\\\\ -------------------------------\\\\ \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-2=7(x-1)\implies y-2=7x-7 \\\\\\ y=7x-5$

Answer 2

F(x)=(x³+1)(3x²-4x+2). Find f ' (x) the derivative of f(x)
f '(x) = (3x²)(3x²-4x+2) + (6x-4)(x³+1), expand:

f '(x) = 15x⁴ - 16x³ + 6x² + 6x - 4
f '(1) = 15(1)⁴ - 16(1)³ + 6(1)² + 6(1) - 4
f '(1) = 7 [7 is the slope "m" at (1,2)]
So the equation of the tangent at (1,2) is:

y = mx + b  
y = 7x + b. To calculate b, plug in (x=1 and y = 2)

2 = 7(1) + b and b = -5. Then the final equation of the tangent at (1,2) is

y = 7x - 5