Question

a. Use definition ( 2 ) ( \(p .\) 135 ) to find the slope of the line tangent to the graph of \(f\) at \(P\). b. Determine an equation of the tangent line at \(P\). $$f(x)=\sqrt{x-1} ; P(2,1)$$

Short answer

Answer: The equation of the tangent line is \(y = \frac{1}{2}(x-2) + 1\).

Step-by-step solution

Find the derivative using the definition

We begin by using the definition of the derivative to find \(f'(x)\), the derivative of the function \(f(x)=\sqrt{x-1}\). The definition for the derivative is: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ So we will first plug in \(f(x)\): $$f'(x) = \lim_{h \to 0} \frac{\sqrt{(x+h)-1} - \sqrt{x-1}}{h}$$ Then, we must simplify this expression and evaluate the limit as \(h\) approaches 0.

Rationalize the numerator

To simplify the expression, we can rationalize the numerator by multiplying the numerator and denominator by the conjugate of the numerator: $$f'(x) = \lim_{h \to 0} \frac{\sqrt{(x+h)-1} - \sqrt{x-1}}{h} \cdot \frac{\sqrt{(x+h)-1} + \sqrt{x-1}}{\sqrt{(x+h)-1} + \sqrt{x-1}}$$ This simplifies to: $$f'(x) = \lim_{h \to 0} \frac{(x+h)-1 - (x-1)}{h(\sqrt{(x+h)-1} + \sqrt{x-1})}$$

Simplify and evaluate the limit

Now, further simplify the expression and evaluate the limit: $$f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{(x+h)-1} + \sqrt{x-1})}$$ Cancel the term \(h\) from the numerator and denominator: $$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{(x+h)-1} + \sqrt{x-1}}$$ Now, we can find the derivative by evaluating the limit as \(h\) approaches 0: $$f'(x) = \frac{1}{\sqrt{x-1} + \sqrt{x-1}}$$ $$f'(x) = \frac{1}{2\sqrt{x-1}}$$

Find the slope of the tangent line at point \(P\)

Now that we have the derivative, which represents the slope of the tangent line at any point, we can find the slope at the given point \(P(2,1)\). We need to substitute \(x = 2\) in the expression for the derivative: $$f'(2) = \frac{1}{2\sqrt{2-1}}$$ $$f'(2) = \frac{1}{2}$$ So the slope of the tangent line at point \(P\) is ½.

Determine the equation of the tangent line using point-slope form

To find the equation of the tangent line, we can use the point-slope form of a linear equation, which is given by: $$y - y_1 = m(x - x_1)$$ Here, \((x_1, y_1)\) represents the point \(P(2,1)\) and \(m\) represents the slope of the tangent line, which we found to be ½. Plugging in these values, we get: $$y - 1 = \frac{1}{2}(x - 2)$$ Now, we can simplify the equation to find the tangent line: $$y = \frac{1}{2}(x-2) + 1$$ The equation of the tangent line is: \(y = \frac{1}{2}(x-2) + 1\).

Key concepts

Derivative

The concept of a derivative is central to calculus and represents the rate at which a function is changing at any given point. For a function \( f(x) \), its derivative, denoted as \( f'(x) \), provides the slope of the tangent line to the function at a given point. The derivative tells us how the output value of the function changes as the input value changes. This is essential for understanding the behavior of functions, predicting future values, and solving various real-world problems.

• The derivative is calculated as the limit of the average rate of change of the function over an interval as the interval becomes infinitesimally small.
• For any function \( f(x) \), the derivative \( f'(x) \) is defined as:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] This formula calculates the slope of the tangent line at a point by considering how the function value changes as \( x \) changes by an infinitesimally small amount \( h \). Understanding derivatives provides a foundation for further studies in calculus, including optimization and motion analysis.

Slope of Tangent Line

The slope of a tangent line at a particular point on a graph of a function is important for understanding the instantaneous rate of change at that particular point. When we refer to the 'tangent line', we mean the straight line that just touches the curve at that specific point without intersecting it over a small interval.

• The slope of the tangent line is precisely the value of the derivative of the function at that point.
• In the mentioned exercise, the function \( f(x) = \sqrt{x-1} \) is given, and the point of tangency is \( P(2,1) \).

To find the slope of the tangent line at \( P \), we evaluate the derivative of \( f(x) \) at \( x=2 \). As calculated, \( f'(2) = \frac{1}{2} \). This means that at \( x=2 \), the function \( f(x) \) is changing at a rate of 0.5 units of output for each unit of input.
Knowing the slope allows us to write the equation for the tangent line, providing information on the linear approximation of the function near that point.

Limit Definition

The limit definition of a derivative is a fundamental concept in calculus. It lays the groundwork for understanding how functions behave as inputs approach particular values, even if they're not specifically defined there.
The limit is defined as the value that a function \( f(x) \) approaches as the input \( x \) gets infinitely close to a particular number. In calculus, we typically express this by writing:

• Limits help in defining the derivative as they allow the calculation of the slope of the tangent line without directly needing a defined point of tangency.
• The limit definition is applied as:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]This definition elegantly captures the idea of a function's instantaneous rate of change, an idea originally introduced by Newton and Leibniz. In practical terms, it enables us to not only find slopes of tangent lines but also solve more complex problems regarding velocity, acceleration, and other rates of change in a wide array of scientific and engineering fields.