Question

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:

\(y = 5x-1\)