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 at (x = 1) is (m_{tangent}=5). Since the slope of the normal (m_{normal}) and the slope of the 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}): (y(1)=3\times1^{2}+\frac{1}{1}=3 + 1=4). So 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 is (y - 4=-\frac{1}{5}(x - 1)). Expand it: (y-4=-\frac{1}{5}x+\frac{1}{5}). Then (y=-\frac{1}{5}x+\frac{1}{5}+4), which simplifies to (y=-\frac{1}{5}x+\frac{21}{5}).

Answer:

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