Problem 73 - Calculus for Biology and Medicine

Question

Find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. \(f(x)=\frac{x+5}{x^{3}}\), at \(x=2\)

Step-by-Step Solution

Answer

The tangent line to the function at \(x = 2\) is \(y = -\frac{19}{16}x + \frac{13}{4}\).

Step 1: Find the Derivative

First, find the derivative of the given function to determine the slope of the tangent line. The function is given as \[ f(x) = \frac{x+5}{x^3} = x^{-2} + 5x^{-3}. \] This can be rewritten using the power rule, \[ f'(x) = -2x^{-3} - 15x^{-4}. \] This expression is equivalent to \[ f'(x) = -\frac{2}{x^3} - \frac{15}{x^4}. \]

Step 2: Evaluate the Derivative at x=2

Next, evaluate the derivative at the point where \(x = 2\) to find the slope of the tangent line. Substitute \(x = 2\) into the derivative:\[ f'(2) = -\frac{2}{2^3} - \frac{15}{2^4} = -\frac{2}{8} - \frac{15}{16}. \] Combine the terms to get a common denominator:\[ f'(2) = -\frac{4}{16} - \frac{15}{16} = -\frac{19}{16}. \] Thus, the slope \(m\) of the tangent line at \(x = 2\) is \(-\frac{19}{16}\).

Step 3: Find the Point on the Function

Determine the y-coordinate of the function at \(x = 2\). Substitute \(x = 2\) into the original function:\[ f(2) = \frac{2+5}{2^3} = \frac{7}{8}. \] The point on the function when \(x = 2\) is \((2, \frac{7}{8})\).

Step 4: Write the Equation of the Tangent Line

Now write the equation of the tangent line in point-slope form, and then convert to slope-intercept form. Using the point \((2, \frac{7}{8})\) and the slope \(-\frac{19}{16}\), the point-slope form is:\[ y - \frac{7}{8} = -\frac{19}{16}(x - 2). \] Distribute and solve for \(y\):\[ y = -\frac{19}{16}x + \frac{19}{8} + \frac{7}{8} = -\frac{19}{16}x + \frac{26}{8} = -\frac{19}{16}x + \frac{13}{4}. \]

Step 5: Conclusion

The equation of the tangent line in slope-intercept form is \[ y = -\frac{19}{16}x + \frac{13}{4}. \]

Key Concepts

Derivative

The derivative of a function provides us with the slope of the tangent line to the curve at any given point. It essentially measures how the function is changing at that particular point. To find the derivative of our function, we start with the expression given: \[ f(x) = \frac{x+5}{x^3} = x^{-2} + 5x^{-3}. \] By applying the power rule of calculus, which states that \(\frac{d}{dx}x^n = nx^{n-1}\), we can compute the derivative as: \[ f'(x) = -2x^{-3} - 15x^{-4}. \] This can also be written as a fraction: \[ f'(x) = -\frac{2}{x^3} - \frac{15}{x^4}. \] This derivative expression allows us to calculate the slope of the tangent line at any point \(x\) on the curve.

Slope-Intercept Form

The slope-intercept form of the equation of a line is a way to represent the linear equation in an easy and useful format. It is written as \(y = mx + b\), where \(m\) is the slope of the line and \(b\) is the y-intercept. In this case:

• \(m\) represents the rate of change at the point of contact on the graph.
• \(b\) indicates where the line crosses the y-axis.

For our tangent line, the slope has been found as \(-\frac{19}{16}\) through derivative evaluation, and the intercept \(b\) was calculated during the conversion from point-slope to slope-intercept form.The final expression for the line is:\[ y = -\frac{19}{16}x + \frac{13}{4}. \] This form is straightforward and useful for quickly assessing a line's steepness and y-intercept.

Point-Slope Form

The point-slope form of a line's equation is particularly handy when you know a specific point on the line and the slope. It is written as \(y - y_1 = m(x - x_1)\), where:

• \((x_1, y_1)\) is a given point on the line.
• \(m\) is the slope of the line.

In our example, we use the point \((2, \frac{7}{8})\) and the slope \(-\frac{19}{16}\). Substituting these into the point-slope form gives:\[ y - \frac{7}{8} = -\frac{19}{16}(x - 2). \] This form makes it simple to generate the line equation directly from a point and a slope, which is critical in finding tangent lines.

Function Evaluation

Function evaluation is the process of substituting a specific value into a function in order to find the corresponding output. This is crucial for finding points on the curve of the function. In our context, we perform function evaluation to determine the y-coordinate at a specific x-value, namely \(x = 2\).

• We substitute \(x = 2\) into the original function \(f(x)\).
• This gives us \[ f(2) = \frac{2 + 5}{2^3} = \frac{7}{8}. \]

By evaluating the function at this point, we obtain the exact coordinates \((2, \frac{7}{8})\) of the point on the curve where the tangent line meets the function. This point is essential for both point-slope form and verifying the tangent line's correctness.