91Ó°ÊÓ

In calculus, the derivative is a fundamental concept that helps us understand the rate at which a function is changing at any given point. Specifically, the derivative provides us the slope of the tangent line to the function at a particular point.
For the function given in the exercise, \(y = \sqrt[3]{x} = x^{1/3}\), we determine the derivative using the power rule. The power rule states that if you have a function \(f(x) = x^n\), the derivative \(f'(x)\) is \(nx^{n-1}\). Applying this rule, we get the derivative:
\[y' = \frac{1}{3}x^{-2/3} = \frac{1}{3}\frac{1}{x^{2/3}}.\]
This expression gives us the slope of the tangent at any point \(x\). To find the exact slope at \(x = 8\), we substitute 8 into the derivative, resulting in \(\frac{1}{12}\). This slope is used for the equation of the tangent line.

The cube root approximation is a technique used to estimate the cube root of a number near a known value. In the exercise, after finding the equation of the tangent line to \(y = \sqrt[3]{x}\) at \(x = 8\), we want to approximate \(\sqrt[3]{9}\).
Since \(x = 8\) is close to \(x = 9\), the tangent line provides a good approximation for \(\sqrt[3]{9}\). In calculus, tangent lines are commonly used to make such approximations because they represent the behavior of the function near the point of tangency. Using the equation:
\[y = \frac{1}{12}x + \frac{2}{3},\]
replace \(x\) with 9 to find \(y\), giving us:
\[y = \frac{9}{12} + \frac{8}{12} = \frac{17}{12}.\]
This result then approximates \(\sqrt[3]{9} \approx 1.417\), providing a practical method to estimate cube roots near known values.

The point-slope form is a useful way to write the equation of a line when you know a point on the line and its slope. The formula is:
\[y - y_1 = m(x - x_1)\]
where \(m\) is the slope, and \((x_1, y_1)\) is the point through which the line passes.
In this exercise, after calculating the derivative, we found the slope of the tangent line at \(x = 8\) to be \(\frac{1}{12}\). We also identified the point of tangency as \((8, 2)\).
Substituting these into the point-slope form, we obtain:
\[y - 2 = \frac{1}{12}(x - 8).\]
This allows us to find the linear equation describing the tangent line, which simplifies to:
\[y = \frac{1}{12}x + \frac{2}{3}.\]
The point-slope form not only helps construct the tangent line equation but also gives insight into how the line behaves around the point \((8, 2)\), a crucial aspect for approximation tasks.