Problem 32 - Calculus Single Variable

Question

Calculate the linearization \(L(x)=\) \(f(0)+f^{\prime}(0) \cdot x\) for the given function \(f\) at \(c=0\) $$ f(x)=x \sqrt{1+x} $$

Step-by-Step Solution

Answer

The linearization of the function at \(c = 0\) is \(L(x) = x\).

Step 1: Differentiate the function

To find the linearization, we need the derivative of the function. The given function is \(f(x) = x \sqrt{1+x}\). Use the product rule for differentiation: if \(u(x) = x\) and \(v(x) = \sqrt{1+x}\), then \(f(x) = u(x)\cdot v(x)\) and \(f'(x) = u'(x)v(x) + u(x)v'(x)\). Calculate \(u'(x)\) and \(v'(x)\), where \(u'(x) = 1\) and \(v'(x) = \frac{1}{2\sqrt{1+x}}\).

Step 2: Apply the product rule

Using the product rule from the previous step, we have: \(f'(x) = (1)\sqrt{1+x} + x\left(\frac{1}{2\sqrt{1+x}}\right)\). Simplify the expression to find \(f'(x)\): \(f'(x) = \sqrt{1+x} + \frac{x}{2\sqrt{1+x}}\).

Step 3: Evaluate the function and its derivative at \(c=0\)

Calculate \(f(0)\) and \(f'(0)\): \(f(0) = 0 \cdot \sqrt{1+0} = 0\). \(f'(0) = \sqrt{1+0} + \frac{0}{2\sqrt{1+0}} = 1\).

Step 4: Write the linearization \(L(x)\)

Using the formula for linearization \(L(x) = f(c) + f'(c) \cdot x\), plug in the values we found: \(L(x) = 0 + 1 \cdot x\). Thus, the linearization is \(L(x) = x\).

Key Concepts

Derivative calculation

Calculating the derivative of a function is essential to determine its rate of change. In this exercise, we need to find the derivative of the function \(f(x) = x \sqrt{1+x}\). Differentiation involves finding the derivative \(f'(x)\) of the function \(f(x)\), which helps in the process of linearization by acting as the slope at a particular point. For functions involving products, such as in our example, using the product rule is necessary. Differentiation shows us how the function behaves at different values, especially around the point we're examining, which in this case is \(c = 0\). Ensuring correct derivative calculation is fundamental for any further steps, like product rule applications or function evaluation.

Product rule

When dealing with a function that is the product of two functions, the product rule is essential. The product rule states that the derivative of a product of two functions \(u(x)\) and \(v(x)\) is given by:

• \(f'(x) = u'(x)v(x) + u(x)v'(x)\)

In our situation, we have the function \(f(x) = x \sqrt{1+x}\), where \(u(x) = x\) and \(v(x) = \sqrt{1+x}\). Therefore:

• \(u'(x) = 1\)
• \(v'(x) = \frac{1}{2\sqrt{1+x}}\)

The product rule allows us to combine these two derivatives into a single expression by plugging them into the formula, resulting in:
\[f'(x) = (1)\sqrt{1+x} + x\left(\frac{1}{2\sqrt{1+x}}\right) = \sqrt{1+x} + \frac{x}{2\sqrt{1+x}}\].
This computation makes it possible to analyze how the combined function behaves near the point of interest.

Function evaluation

After calculating the derivative, the next step is to evaluate the function and its derivative at a specific point. For this exercise, we are calculating at \(c = 0\). Evaluating a function means finding its value at a given point.

• For \(f(x) = x \sqrt{1+x}\), at \(x = 0\), the function evaluates to \(f(0) = 0 \cdot \sqrt{1+0} = 0\).
• For the derivative, \(f'(x) = \sqrt{1+x} + \frac{x}{2\sqrt{1+x}}\), we evaluate it at \(x = 0\) to get \(f'(0) = \sqrt{1+0} + \frac{0}{2\sqrt{1+0}} = 1\).

Evaluating these values helps in constructing the linear approximation or the linearization of \(f(x)\). By using the formula for linearization \(L(x) = f(c) + f'(c) \cdot x\), we find that \(L(x) = 0 + 1 \cdot x = x\). This means around \(c = 0\), \(f(x)\) behaves approximately like the function \(x\), helping in estimating the function's behavior with simple linear expressions. This simplified expression can be incredibly useful for approximations and quick predictions about the function's behavior.