Problem 25 - Essential Calculus

Question

In Exercises 25 and \(26,\) verify the tangent line approximation of the function at the given point. Then use a graphing utility to graph the function and its approximation in the same viewing window. $$ \begin{array}{llll} \text { Function } & & \text { Approximation } & \text { Point } \\ & & & \\ f(x)=\sqrt{x} & & y=\frac{1}{2}+\frac{x}{2} & & (1,1) \end{array} $$

Step-by-Step Solution

Answer

The tangent line approximation \(y = \frac{1}{2}+\frac{x}{2}\) to the function \(f(x)=\sqrt{x}\) at the point (1,1) is confirmed to be correct.

Step 1: Find the original function's derivative

Taking the derivative of the function \(f(x) = \sqrt{x}\) by using power rule yields \(f'(x) = \frac{1}{2\sqrt{x}}\) or \(\frac{1}{2x^{\frac{1}{2}}}\).

Step 2: Evaluate the derivative at the point

Substituting \(x = 1\) into the derivative, we find \(f'(1) = \frac{1}{2}\). This is the slope of the tangent line at the point (1,1).

Step 3: Establish the linear approximation

The linear approximation or tangent line at the point (1,1) would be given by \(y = f(1) + f'(1)(x-1)\), substituting for \(f(1) = \sqrt{1} = 1\) and \(f'(1) = \frac{1}{2}\), we get: \(y = 1 + \frac{1}{2}(x-1)\), which simplifies to \(y = \frac{1}{2} + \frac{x}{2}\)

Step 4: Compare the approximation

Looking at the approximation found, it is actually the same as the one we were given originally (\(y = \frac{1}{2}+\frac{x}{2}\)), thus confirming that the given function indeed represents the tangent line to the original function at point (1,1).

Step 5: Graphing

Finally, graphing the original function and the tangent line using any graphing utility would show how well the approximation fits the function near the point (1,1).

Key Concepts

Derivative

Understanding derivatives is crucial when dealing with tangent line approximations. The derivative of a function represents the rate at which the function's value changes as the input changes. This rate of change is essentially the slope of the tangent line to the function at any given point.

In the context of our exercise, we have the function \( f(x) = \sqrt{x} \). The derivative, denoted as \( f'(x) \), is found using the power rule. For \( f(x) = x^{1/2} \), the power rule tells us that the derivative is \( f'(x) = \frac{1}{2}x^{-1/2} \), or more simply, \( f'(x) = \frac{1}{2\sqrt{x}} \). This formula allows us to find how steeply the curve rises or falls at any point \( x \).

A tangent line, which is the best linear approximation of a function near a point, has a slope equal to the derivative of the function at that point. Therefore, by calculating the derivative at \( x = 1 \), you get \( f'(1) = \frac{1}{2} \), establishing that the slope of our tangent line at \((1,1)\) is \( \frac{1}{2} \).

Graphing Utility

A graphing utility is a tool that allows you to visually represent mathematical functions and their approximations. This can be immensely helpful in understanding how a function behaves near a specific point and how well its tangent line approximates it.

Using software like Desmos or a graphing calculator, you can input the original function \( f(x) = \sqrt{x} \) and its tangent line approximation \( y = \frac{1}{2} + \frac{x}{2} \).
Such a visual can offer a clear insight into whether the linear approximation is valid by showing the tangent line closely hugging the curve at \( (1, 1) \).

This technique directly complements the analytical approach, bringing a graphical clarity to the mathematical approximation. It also aids in verifying your calculations, as the perfect alignment of the curve and the tangent line at the point confirms that the approximation was correctly derived.

Linear Approximation

Linear approximation, or tangent line approximation, is a method to approximate the value of a function near a given point by using the equation of its tangent line. It is particularly useful when a function is too complex to compute directly.

The linear approximation formula is built with:

• The value of the function at the point, \( f(a) \)
• The derivative of the function at the point, \( f'(a) \)
• The variable change from the point, \( (x - a) \)

Applying these, the formula becomes \( y = f(a) + f'(a)(x-a) \).

For our function \( f(x) = \sqrt{x} \) at \( x = 1 \), we plug in values \( f(1) = 1 \) and \( f'(1) = \frac{1}{2} \).
Thus, our linear approximation equation becomes \( y = 1 + \frac{1}{2}(x - 1) \), which simplifies to \( y = \frac{1}{2} + \frac{x}{2} \). This is the equation for the tangent line, serving as a linear approximation of the function near the point \((1, 1)\).

This linear model is straightforward and provides a good approximation for small intervals around a specific point, making it invaluable in calculus for estimations and solving problems involving rates of change.