Problem 73 - Calculus for Biology and Medicine

Question

Find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. $$ f(x)=\frac{2 x-5}{x^{3}}, \text { at } x=2 \quad \text { 74. } f(x)=\sqrt{x}\left(x^{3}-1\right), \text { at } x=1 $$

Step-by-Step Solution

Answer

For function 1: y = \(\frac{7}{16}x - 1\). For function 2: y = 3x - 3.

Step 1: Identify the Problem

We need to find the equation of the tangent line in slope-intercept form (\(y = mx + b\)) for the functions at given points. The tangent line necessitates finding the derivative of the function to get the slope \(m\) at the specified \(x\) values and using function values to solve for \(b\).

Step 2: Calculate Derivative of First Function

For \(f(x) = \frac{2x-5}{x^3}\), apply the quotient rule: If \(u = 2x - 5\) and \(v = x^3\), then \(f'(x) = \frac{u'v - uv'}{v^2}\). Compute \(u' = 2\) and \(v' = 3x^2\). Substitute to find \(f'(x) = \frac{2x^3 - (2x - 5) \cdot 3x^2}{x^6}\).

Step 3: Simplify the Derivative of First Function

Simplify \(f'(x) = \frac{2x^3 - 6x^3 + 15x^2}{x^6} = \frac{-4x^3 + 15x^2}{x^6} = -\frac{4}{x^3} + \frac{15}{x^4}\).

Step 4: Evaluate Derivative at x=2 for First Function

Substitute \(x = 2\) into \(f'(x)\). Calculate \(-\frac{4}{2^3} + \frac{15}{2^4} = -\frac{4}{8} + \frac{15}{16} = -\frac{1}{2} + \frac{15}{16} = \frac{16}{16} - \frac{8}{16} = \frac{7}{16}\). So, the slope \(m = \frac{7}{16}\).

Step 5: Find y-value of First Function at x=2

Substitute \(x = 2\) into \(f(x)\): \(f(2) = \frac{2(2) - 5}{2^3} = \frac{4 - 5}{8} = -\frac{1}{8}\). The point is \((2, -\frac{1}{8})\).

Step 6: Equation of Tangent Line for First Function

Using point-slope form \(y - y_1 = m(x - x_1)\) with \(m = \frac{7}{16}\), \(x_1 = 2\), and \(y_1 = -\frac{1}{8}\), convert to slope-intercept form: \(y + \frac{1}{8} = \frac{7}{16}(x - 2)\). Simplify \(y = \frac{7}{16}x - \frac{7}{8} - \frac{1}{8} = \frac{7}{16}x - 1\). So, \(y = \frac{7}{16}x - 1\).

Step 7: Calculate Derivative of Second Function

For \(f(x) = \sqrt{x}(x^3-1)\), use the product rule: If \(u = \sqrt{x}\) and \(v = x^3 - 1\), \(f'(x) = u'v + uv'\). Compute \(u' = \frac{1}{2\sqrt{x}}\) and \(v' = 3x^2\). Substitute: \(f'(x) = \frac{1}{2\sqrt{x}}(x^3 - 1) + \sqrt{x}(3x^2)\).

Step 8: Simplify the Derivative of Second Function

\(f'(x) = \frac{x^3 - 1}{2\sqrt{x}} + 3x^{5/2}\). Simplify terms to a single expression if needed.

Step 9: Evaluate Derivative at x=1 for Second Function

Substitute \(x = 1\) into \(f'(x)\). Calculate \(\frac{1^3 - 1}{2\cdot1} + 3\times1^{5/2} = 0 + 3 = 3\). So, the slope \(m = 3\).

Step 10: Find y-value of Second Function at x=1

Substitute \(x = 1\) into \(f(x)\): \(f(1) = \sqrt{1}(1^3 - 1) = 0\). The point is \((1, 0)\).

Step 11: Equation of Tangent Line for Second Function

Using point-slope form \(y - y_1 = m(x - x_1)\) with \(m = 3\), \(x_1 = 1\), and \(y_1 = 0\), convert to slope-intercept form: \(y = 3(x - 1) = 3x - 3\).

Key Concepts

Quotient Rule

The **quotient rule** helps us find the derivative of a function that is a ratio of two other functions. This is critical when your function resembles a fraction, as in the case of

• \( f(x) = \frac{2x - 5}{x^3} \).

To differentiate such a function, define the numerator as \( u \) and the denominator as \( v \). In our example:

• \( u = 2x - 5 \)
• \( v = x^3 \)

The derivative of a quotient, \( \frac{u}{v} \), is given by the formula

• \( f'(x) = \frac{u'v - uv'}{v^2} \)

Where:

• \( u' = 2 \) is the derivative of \( u \)
• \( v' = 3x^2 \) is the derivative of \( v \)

When you apply these derivatives to the formula, it allows you to understand how the quotient rule provides an effective way to differentiate ratios of functions easily. Applying this rule requires careful simplification to reach the correct derivative.

Product Rule

The **product rule** is employed when differentiating a product of two functions, such as

• \( f(x) = \sqrt{x}(x^3 - 1) \).

To apply this rule, treat each part of the product as a separate function:

• \( u = \sqrt{x} \)
• \( v = x^3 - 1 \)

The product rule states that the derivative of \( uv \) is:

• \( f'(x) = u'v + uv' \)

Here:

• \( u' = \frac{1}{2\sqrt{x}} \) is the derivative of \( u \), observing the chain rule for square roots
• \( v' = 3x^2 \) is the power rule applied to \( v \)

Substituting these derivatives back into the product rule formula allows for a precise derivative to be found. Be sure to simplify the resulting expression to make subsequent evaluations, like plugging in specific \( x \) values, much easier.

Derivative Evaluation

After finding the derivatives using rules like the quotient or product rules, **derivative evaluation** involves computing the value of these derivatives at a specified \( x \) value to find the slope of the tangent line. This slope is essentially how steep or flat the function is at that precise point. Let's go through the steps:1. **Substitute the given \( x \) value into the derivative.** - For the first function, with \( x = 2 \), evaluate \(-\frac{4}{2^3} + \frac{15}{2^4} = \frac{7}{16}\), meaning the slope of the tangent line at \( x = 2 \) is \( \frac{7}{16}\).2. **Use the point to find the complete equation of the tangent line.** Singling out the point on the original curve, - At \( x = 2 \), use \( f(2) = -\frac{1}{8} \) - showing our point is \((2, -\frac{1}{8})\).3. **Form the tangent line equation**: Using point-slope form, solve for the slope-intercept form: - Substitute: \( y - (-\frac{1}{8}) = \frac{7}{16}(x - 2) \) yields the equation \( y = \frac{7}{16}x - 1 \).This approach gives the complete equation of the tangent line, summarizing the function’s behavior at that particular \( x \). Understanding how to evaluate derivatives at specific points is crucial for accurately describing function properties, especially when visualizing graphs.