Problem 38 - Single Variable Calculus

Question

(a) find an equation of the tangent line to he graph of the function at the indicated point, and (b) use a graphing utility to plot the graph of the function and the tangent line on the same screen. \(f(x)=\sqrt{x}+\frac{1}{\sqrt{x}} ;\left(4, \frac{5}{2}\right)\)

Step-by-Step Solution

Answer

The equation of the tangent line to the graph of the function \(f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}\) at the point \((4, \frac{5}{2})\) is \(y=\frac{3}{8}(x-4)+\frac{5}{2}\).

Step 1: Find the derivative of the function

In order to find the equation of the tangent line, we first need to find the derivative of the function. The function is \(f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}\). We can rewrite this function as \(f(x)=x^{\frac{1}{2}}+x^{-\frac{1}{2}}\). Now, to find the derivative, we will use the power rule: \(\frac{d}{dx}[x^n]=nx^{n-1}\). Applying the power rule to our function, we get: \(f'(x)= \frac{1}{2}x^{\frac{1}{2}-1}-\frac{1}{2}x^{-\frac{1}{2}-1}\) Which simplifies to: \(f'(x)= \frac{1}{2}x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{3}{2}}\)

Step 2: Find the slope of the tangent line

Now that we have the derivative of the function, we need to find the slope of the tangent line at the point \((4, \frac{5}{2})\). To do this, we plug the x-coordinate, 4, into our derivative: \(f'(4)= \frac{1}{2}(4)^{-\frac{1}{2}}-\frac{1}{2}(4)^{-\frac{3}{2}}\) This simplifies to: \(f'(4)= \frac{1}{2}-\frac{1}{8}\) So the slope of the tangent line at the given point is \(\frac{3}{8}\).

Step 3: Find the equation of the tangent line

Now that we have the slope and a point on the tangent line, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is \(y-y_1=m(x-x_1)\), where \((x_1, y_1)\) is a point on the line and m is the slope. Plugging in our point and slope, we get: \(y-\frac{5}{2}=\frac{3}{8}(x-4)\) Now we can solve for y: \(y=\frac{3}{8}(x-4)+\frac{5}{2}\) So, the equation of the tangent line is \(y=\frac{3}{8}(x-4)+\frac{5}{2}\).

Step 4: Graph the function and the tangent line

Finally, we will use a graphing utility to plot the graph of the function and the tangent line on the same screen. The function is \(f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}\), and the tangent line equation is \(y=\frac{3}{8}(x-4)+\frac{5}{2}\). When graphing these together, you will see that the tangent line touches the curve of the function at the given point \((4, \frac{5}{2})\).

Key Concepts

Derivative

A derivative represents the rate at which a function is changing at any given point. It is a fundamental concept in calculus, crucial for understanding how functions behave. When you find the derivative of a function, you're essentially finding the slope of the tangent line to the curve at any point on the graph. The derivative is usually denoted by \( f'(x) \) or \( \frac{df}{dx} \). Derivatives allow us to find acceleration, optimize situations, and analyze functional behavior. Calculating derivatives involves learning specific rules, one of which is the Power Rule.

Power Rule

The Power Rule is a straightforward and efficient method for finding the derivative of expressions in the form of \( x^n \), where \( n \) is any real number. It states that the derivative of \( x^n \) is \( nx^{n-1} \). This method simplifies the process of differentiation, making it quicker and more intuitive. For example, if you start with \( f(x) = x^{1/2} + x^{-1/2} \), applying the power rule gives us \( f'(x) = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} \). This step is crucial for solving problems involving tangent lines because it allows you to calculate the exact slope at any point on the curve, which leads us to the point-slope form.

Point-Slope Form

The point-slope form is a mathematical formula used to find the equation of a line when you know a point on the line and the line's slope. This form is written as \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point and \( m \) is the slope. Once you have the slope of the tangent line from the derivative and the specific point it touches the curve, substituting these into the point-slope form will yield the tangent line's equation. For instance, using point \((4, \frac{5}{2})\) and slope \( \frac{3}{8} \) gives \( y - \frac{5}{2} = \frac{3}{8}(x - 4) \). This formula is instrumental when calculating exact tangent lines on curves.

Graphing Utility

A graphing utility, such as a graphing calculator or software, is a useful tool for visualizing mathematical concepts. It allows you to plot equations and see their graphical representation on the same screen. For instance, graphing the function \( f(x) = \sqrt{x} + \frac{1}{\sqrt{x}} \) alongside its tangent line \( y = \frac{3}{8}(x - 4) + \frac{5}{2} \), helps in understanding the exact point where they touch. This visual approach can make abstract concepts more concrete, allowing for better comprehension of how the function's rate of change is depicted graphically. Graphing utilities provide an interactive way to explore and confirm the results of calculus problems, enhancing the learning experience.