Question

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:

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