The problem asks to find the equation of the tangent line to the graph of the function $f(x) = \frac{2x^3 - 1}{x^2}$ at the point $(1,1)$.

Steps to solve:

1. Simplify the function: $f(x) = \frac{2x^3 - 1}{x^2} = 2x - \frac{1}{x^2}.$

2. Find the derivative $f'(x)$ to determine the slope of the tangent line: We use the sum and power rules to differentiate: $f'(x) = \frac{d}{dx}(2x) - \frac{d}{dx}\left(\frac{1}{x^2}\right).$

   • The derivative of $2x$ is $2$.
   • The derivative of $\frac{1}{x^2}$ is $-2x^{-3}$, which simplifies to $\frac{2}{x^3}$.

   Therefore, $f'(x) = 2 + \frac{2}{x^3}.$

3. Evaluate $f'(x)$ at $x = 1$ to get the slope of the tangent line at the point $(1,1)$: $f'(1) = 2 + \frac{2}{1^3} = 2 + 2 = 4.$ So, the slope of the tangent line is $4$.

4. Use the point-slope form of the equation of a line to find the tangent line: The point-slope form is: $y - y_1 = m(x - x_1),$ where $m$ is the slope and $(x_1, y_1)$ is the point. Here, $m = 4$ and $(x_1, y_1) = (1, 1)$.

   Substituting into the point-slope form: $y - 1 = 4(x - 1).$

5. Simplify the equation: $y - 1 = 4x - 4,$ $y = 4x - 3.$

Final Answer:

The equation of the tangent line to the function at $(1, 1)$ is: $y = 4x - 3.$

Would you like further details or have any questions?

Here are 5 related questions for further practice:

1. How do you find the derivative of more complicated rational functions?
2. What is the geometric meaning of a tangent line to a curve?
3. How do you find the normal line to a curve at a given point?
4. Can you find the equation of the tangent line to a parabola at any point?
5. How would the process differ if the function was implicitly defined?

Tip: When finding tangent lines, always evaluate the derivative at the given point to get the slope before using the point-slope form of the equation of a line.