Answer:
To find the local linearization of the equation \(xy = 8\) at \(x = 2.5\), we'll use the concept of linear approximation, which involves finding the equation of the tangent line to the curve at the given point. The equation of a line in point-slope form is given by:
\[y - y_1 = m(x - x_1)\]
Where:
- \(y\) and \(x\) are the variables in question.
- \((x_1, y_1)\) is the point we want to approximate around.
- \(m\) is the slope of the tangent line.
Let's first find the slope, \(m\), at the point \((2.5, y_1)\). To do this, we'll need to find both \(\frac{dy}{dx}\) and evaluate the equation at \(x = 2.5\).
Given the equation \(xy = 8\), we can implicitly differentiate it with respect to \(x\):
\[\frac{d(xy)}{dx} = \frac{d(8)}{dx}\]
Using the product rule for differentiation, we get:
\[x\frac{dy}{dx} + y\frac{dx}{dx} = 0\]
Since \(\frac{dx}{dx} = 1\), we can simplify to:
\[x\frac{dy}{dx} + y = 0\]
Now, we need to evaluate this at \(x = 2.5\). Plugging in \(x = 2.5\) gives us:
\[2.5\frac{dy}{dx} + y_1 = 0\]
Now, we need to solve for \(\frac{dy}{dx}\), the slope of the tangent line at \(x = 2.5\):
\[2.5\frac{dy}{dx} = -y_1\]
\[\frac{dy}{dx} = -\frac{y_1}{2.5}\]
Next, we'll evaluate the original equation \(xy = 8\) at \(x = 2.5\) to find \(y_1\):
\[2.5y_1 = 8\]
Now, solve for \(y_1\):
\[y_1 = \frac{8}{2.5} = 3.2\]
So, the point \((2.5, y_1)\) is \((2.5, 3.2)\), and we've already found the slope \(\frac{dy}{dx} = -\frac{3.2}{2.5} = -1.28\).
Now, we can write the equation of the tangent line using point-slope form:
\[y - 3.2 = -1.28(x - 2.5)\]
Simplify and rearrange:
\[y - 3.2 = -1.28x + 3.2\]
Now, add \(3.2\) to both sides:
\[y = -1.28x + 6.4\]
So, the local linearization of the curve \(xy = 8\) at \(x = 2.5\) is \(y = -1.28x + 6.4\).
Explanation: