find the equation of the tangent 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) is (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). So the slope (m) of the tangent line at (x = 1) is (5).

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

Substitute (x = 1) into (y = 3x^{2}+\frac{1}{x}), we get (y(1)=3\times1^{2}+\frac{1}{1}=3 + 1=4). So the point ((x_0,y_0)=(1,4)) lies on the curve and the tangent line.

Step4: Use the point - slope form of a line

The point - slope form of a line is (y - y_0=m(x - x_0)). Substituting (m = 5), (x_0 = 1) and (y_0 = 4) into it, we have (y-4=5(x - 1)). Expand the right - hand side: (y-4=5x-5). Add (4) to both sides to get the equation of the tangent line: (y=5x-1).

Answer:

(y = 5x-1)