Problem 72 - Calculus for Biology and Medicine

Question

Find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. $$ f(x)=\frac{3}{x}-\frac{4}{\sqrt{x}}+\frac{2}{x^{2}}, \text { at } x=1 $$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = -5x + 6 \).

Step 1: Differentiate the Function

First, find the derivative of the function \( f(x) = \frac{3}{x} - \frac{4}{\sqrt{x}} + \frac{2}{x^2} \) to get \( f'(x) \). This involves applying the power rule and chain rule where necessary.Rewrite the function:\[ f(x) = 3x^{-1} - 4x^{-1/2} + 2x^{-2} \]Differentiate each term:\[ f'(x) = -3x^{-2} + 2x^{-3/2} - 4x^{-3} \]

Step 2: Evaluate the Derivative at x = 1

Substitute \( x = 1 \) into \( f'(x) \) to find the slope of the tangent line at \( x = 1 \).\[ f'(1) = -3(1)^{-2} + 2(1)^{-3/2} - 4(1)^{-3} \]Calculate the values:\[ = -3 + 2 - 4 = -5 \]

Step 3: Find the Function Value at x = 1

Substitute \( x = 1 \) into the original function \( f(x) \) to find the corresponding \( y \)-value for the point of tangency.\[ f(1) = \frac{3}{1} - \frac{4}{\sqrt{1}} + \frac{2}{1^2} \]Calculate the values:\[ = 3 - 4 + 2 = 1 \]

Step 4: Write the Equation of the Tangent Line

Now use the point-slope form of a line to find the equation of the tangent line, using the point \((1, 1)\) and slope \(-5\).Point-slope form is \( y - y_1 = m(x - x_1) \).Substitute in \( (x_1, y_1) = (1, 1) \) and \( m = -5 \):\[ y - 1 = -5(x - 1) \]Simplify to slope-intercept form (\( y = mx + b \)):\[ y = -5x + 5 + 1 \]\[ y = -5x + 6 \]

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at exactly one point, without crossing it. At this point, the tangent line and the curve share the same slope. This slope tells you how steep the line is at that specific point on the curve.

Understanding tangent lines is crucial because they approximate how the function behaves at that point. If you zoom in very closely on a curve at the point where a tangent line touches, the curve starts to look almost like a straight line. This is why tangent lines are used to make predictions and analyze behavior locally along the curve.

When you're asked to find the tangent line for a function at a specific point, you are essentially finding a line that shows the instant rate of change of the function at that point. This leads us to the concept of differentiation.

Differentiation

Differentiation is a process in calculus that finds the rate at which a function is changing at any given point. It allows us to find the derivative of a function, which gives us the slope of the tangent line at any point along the curve.

To differentiate a function, you apply rules like the power rule, product rule, quotient rule, or chain rule, depending on the form of the function. In the exercise to find the tangent line for the function \(f(x) = \frac{3}{x} - \frac{4}{\sqrt{x}} + \frac{2}{x^{2}}\), we rewrite each term using exponents to make differentiation more straightforward:

• \(3x^{-1}\) differentiates to \(-3x^{-2}\).
• \(-4x^{-1/2}\) differentiates to \(2x^{-3/2}\).
• \(2x^{-2}\) differentiates to \(-4x^{-3}\).

By substituting \(x = 1\) into the derivative \(f'(x) = -3x^{-2} + 2x^{-3/2} - 4x^{-3}\), we found the slope of the tangent line at that point. Differentiation is used to identify how steep the curve is at any given point, which is essential for constructing an accurate tangent line.

Slope-Intercept Form

The slope-intercept form is a way to write the equation of a straight line. It is written as \(y = mx + b\), where \(m\) represents the slope (or steepness) of the line, and \(b\) is the y-intercept (the point where the line crosses the y-axis).

In our exercise, once we have the slope of the tangent line, which is \(-5\), we can use the point-slope form of the equation, which is \(y - y_1 = m(x - x_1)\), to find the exact equation of our tangent line through the point \((1,1)\).

Substituting the values into the point-slope form:

• \(y - 1 = -5(x - 1)\)

We can simplify this equation to reach the familiar slope-intercept form:

• \(y = -5x + 6\)

This gives us a complete picture of the line's behavior and position concerning the original curve. Using this form, you can quickly understand the direction and intensity of the line's slope and its starting position on the graph.