Problem 20 - Precalculus with Limits

Question

In Exercises 17-22, find a formula for the slope of the graph of \(f\) at the point \((x, f(x))\). Then use it to find the slope at the two given points. \(f(x) = \dfrac{1}{x+2}\) (a) \((0, \frac{1}{2})\) (b) \((-1, 1)\)

Step-by-Step Solution

Answer

The slopes of the graph of the function at the points (0, 1/2) and (-1, 1) are -1/4 and -1 respectively.

Step 1 - Differentiate the Function

The derivative of the function \( f(x) = \dfrac{1}{x+2} \) is calculated as follows: The function is in quotient form, so you'll use the quotient rule for differentiation, which states that the derivative of \( \frac{u}{v} \) where \( u \) and \( v \) are functions of \( x \), is \( \frac{vu' - uv'}{v^{2}} \). In this case, \( u = 1 \) and \( v = x + 2 \). The derivatives of \( u \) and \( v \), denoted as \( u' \) and \( v' \), are \( u' = 0 \) and \( v' = 1 \) respectively. Plugging into the formula gives \( f'(x) = \frac{(x+2) \cdot 0 - 1 \cdot 1}{(x + 2)^{2}} \) = \( -\frac{1}{(x+2)^{2}} \). This is the formula for the slope at any point on the graph of the function.

Step 2 - Calculate the slope at (0, 1/2)

Substitute \( x = 0 \) into the formula obtained in the first step to calculate the slope at the point (0, 1/2). Thus, \( f'(0) = -\frac{1}{(0+2)^{2}} = -\frac{1}{4} \).

Step 3 - Calculate the slope at (-1, 1)

Again, substitute \( x = -1 \) into the formula obtained in step one to calculate the slope at the point (-1, 1). Thus, \( f'(-1) = -\frac{1}{(-1+2)^{2}} = -1 \).

Key Concepts

Derivative of a Function

Understanding the derivative of a function is a crucial element in calculus, often interpreted as the instantaneous rate of change or the slope of the tangent line to a curve at a given point. For a function like \(f(x)\), the derivative, denoted as \(f'(x)\), provides us with a powerful tool to analyze various aspects of the function's graph, such as its increasing or decreasing behavior and points of inflection.

In a geometric sense, if you were to zoom in infinitely close on a single point on the graph of a continuous function, the curve would start to resemble a straight line. The slope of this 'almost' straight line is what the derivative at that point represents. When finding the derivative of a single term like \(1/(x+2)\), the function tells us how steeply the graph inclines or declines as \(x\) changes.

Quotient Rule for Differentiation

When dealing with a function that is a ratio of two other functions, we rely on the quotient rule for differentiation to find its derivative. This rule states that if we have a function \(f(x) = u/v\), where both \(u\) and \(v\) are functions of \(x\), the derivative \(f'(x)\) is given by \(\frac{vu' - uv'}{v^2}\).

Take the function \(f(x) = 1/(x+2)\) as an example. Here, \(u = 1\) which is a constant, and \(v = x + 2\) which is a linear function. Their derivatives \(u'\) and \(v'\) are 0 (since the derivative of a constant is 0) and 1 (since the derivative of \(x\) is 1), respectively. Applying the quotient rule, the slope-formula for any given point on the graph \(f'(x)\) simplifies to \( -1/(x+2)^2\).

Calculating Slope at a Point

Calculating the slope at a specific point on the curve of a function involves evaluating the derivative at that point. This gives us insight into how the function is behaving at that precise location on the graph. If \(f'(x)\) is positive, the function is increasing; if it's negative, the function is decreasing.

With our given function \(f(x)=1/(x+2)\), after finding the derivative, we calculate the slope at a specific point by substituting the value of \(x\) into our derivative formula. For instance, to find the slope at \(x=0\), we plug \(0\) into \(f'(x)\) to get \(f'(0)=-1/4\). This negative value signifies that the graph slopes downward at \(x=0\). Similarly, at \(x=-1\), \(f'(-1)=-1\), which again indicates a downward slope but steeper than at \(x=0\).

Differentiation Steps

The process of differentiation follows systematic steps to find the slope formula for a function's graph. Initially, identify the type of function you have (e.g., polynomial, quotient, product) in order to apply the appropriate rule - such as quotient rule in this case. Then, differentiate the numerator and the denominator separately as per the basics of differentiation.

To ensure accuracy, carefully apply the formula for the rule being used, and simplify the result by combining like terms and performing algebraic operations. Regularly reviewing and practicing these steps can significantly enhance one's ability to differentiate functions quickly and correctly. For complex functions, breaking them down into simpler parts can often make the process more manageable.