find the equation of the normal to the graph of $y = 3x^{2}+\frac{1}{x}$ at $x = 1$.

Explanation:

Step1: Find the derivative of the function

The function is (y = 3x^{2}+\frac{1}{x}=3x^{2}+x^{- 1}). Using the power rule ((x^{n})^\prime=nx^{n - 1}), the derivative (y^\prime=6x-x^{-2}=6x-\frac{1}{x^{2}}).

Step2: Evaluate the derivative at (x = 1)

Substitute (x = 1) into (y^\prime): (y^\prime(1)=6\times1-\frac{1}{1^{2}}=6 - 1=5). The slope of the tangent line at (x = 1) is (m_{tangent}=5). Since the slope of the normal line (m_{normal}) and the slope of the tangent line (m_{tangent}) satisfy (m_{normal}\times m_{tangent}=-1), then (m_{normal}=-\frac{1}{5}).

Step3: Find the (y) - coordinate when (x = 1)

Substitute (x = 1) into (y = 3x^{2}+\frac{1}{x}), we get (y=3\times1^{2}+\frac{1}{1}=3 + 1=4). The point on the curve is ((1,4)).

Step4: Use the point - slope form (y - y_{1}=m(x - x_{1}))

Here (x_{1}=1,y_{1}=4,m =-\frac{1}{5}). The equation of the line is (y - 4=-\frac{1}{5}(x - 1)). Expand it: (y-4=-\frac{1}{5}x+\frac{1}{5}). (y=-\frac{1}{5}x+\frac{1}{5}+4). (y=-\frac{1}{5}x+\frac{1 + 20}{5}).

Answer:

(y=-\frac{1}{5}x+\frac{21}{5})