Problem 16
Question
a) Graph the function. b) Draw tangent lines to the graph at points whose \(x\) -coordinates are \(-2,0,\) and 1 c) Find \(f^{\prime}(x)\) by determining \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\). d) Find \(f^{\prime}(-2), f^{\prime}(0),\) and \(f^{\prime}(1) .\) These slopes should match those of the lines you drew in part (b). $$f(x)=\frac{2}{x}$$
Step-by-Step Solution
Verified Answer
Graph \( f(x) = \frac{2}{x} \), find tangent slopes at \(-2, 0, 1\); they are \(-\frac{1}{2},\) undef., and \(-2\).
1Step 1: Graph the function
To graph the function \( f(x) = \frac{2}{x} \), plot points using selected \(x\)-values to ensure you capture the behavior around the vertical asymptote at \(x = 0\) and the hyperbolic shape. Remember that the function is undefined at \(x = 0\) and exhibits two branches in the first and third quadrants.
2Step 2: Draw tangent lines
Identify the points on the curve where \(x = -2, 0,\) and \(1\). Draw tangent lines at these points by estimating the slope visually or using a graphing tool. The lines should touch the graph at exactly one point at the given \(x\)-values.
3Step 3: Use the definition of a derivative
The derivative \(f'(x)\) of a function \(f(x)\) can be found using the limit definition: \[f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}.\] Substituting \(f(x) = \frac{2}{x}\), compute: \[f'(x) = \lim_{h \rightarrow 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h}.\] Simplify the expression to find the derivative.
4Step 4: Simplify to find the derivative
Continuing from the previous step, simplify: \[f'(x) = \lim_{h \rightarrow 0} \frac{\frac{2(x) - 2(x+h)}{x(x+h)}}{h} = \lim_{h \rightarrow 0} \frac{-2h}{hx(x+h)}.\] Cancel \(h\) from the numerator and denominator to get: \[f'(x) = \lim_{h \rightarrow 0} \frac{-2}{x(x+h)}.\] As \(h \rightarrow 0\), the limit simplifies to: \[f'(x) = -\frac{2}{x^2}.\]
5Step 5: Calculate specific derivative values
Substitute \(x = -2, 0,\) and \(1\) into \(f'(x) = -\frac{2}{x^2}\) to find \(f'(-2), f'(0),\) and \(f'(1)\). For \(x = -2\), \[f'(-2) = -\frac{2}{(-2)^2} = -\frac{2}{4} = -\frac{1}{2}.\] For \(x = 0\), \(f'(0)\) is undefined as the original function is not defined at \(x = 0\).For \(x = 1\), \[f'(1) = -\frac{2}{(1)^2} = -2.\] Note that \(f'(0)\) cannot be found graphically either because the tangent line does not exist at \(x = 0\).
6Step 6: Match the slopes with tangent lines
Check that the slopes of the tangent lines drawn in Part (b) match the derivative values calculated in Step 5. Ensure that at \(x = -2\), the slope is \(-\frac{1}{2}\), and at \(x = 1\), the slope is \(-2\). The slope at \(x = 0\) cannot be matched as both the function and the derivative are undefined there.
Key Concepts
Function GraphingTangent LinesLimit Definition of DerivativeSlope of Tangent Line
Function Graphing
In calculus, graphing a function helps visualize its behavior, especially around points of interest like asymptotes or regions where the function is undefined. The function given, \( f(x) = \frac{2}{x} \), is a simple rational function. This function is undefined at \( x = 0 \) as division by zero is not permissible. When graphing, you would typically plot key points to capture the function's behavior accurately. Here:
- The graph has a vertical asymptote at \( x = 0 \), meaning the graph approaches infinity as \( x \) approaches zero from either side.
- The function has two hyperbolic branches, located in the first and third quadrants of the Cartesian plane, indicating where \( x \) takes positive and negative values respectively.
- As \( x \) moves away from zero, towards positive or negative infinity, the function's value approaches zero, showing the horizontal asymptote along the \( y \)-axis.
Tangent Lines
A tangent line touches the curve of a function at precisely one point, providing the instant slope or direction at that point on the graph. Drawing tangent lines can help illustrate how a function behaves locally.
- For our function, \( f(x) = \frac{2}{x} \), tangent lines are drawn at \( x = -2, 0, \) and \( 1 \).
- At \( x = -2 \) and \( x = 1 \), you can visually estimate the slope by drawing a line that barely skims the curve at that exact point and extends outwards following the curve's direction.
- At \( x = 0 \), the function is not defined, and therefore, no tangent line exists.
Limit Definition of Derivative
The limit definition of a derivative is foundational in calculus. It expresses the derivative at a point in terms of limits, which helps describe how the function changes at a very small scale.For any function \( f(x) \), the typical formula for the derivative is:\[ f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \]This formula gives us the slope of the tangent line or the rate of change of the function at a point \( x \).By substituting \( f(x) = \frac{2}{x} \) into this formula:
- You compute \( \lim_{h \rightarrow 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h} \) and simplify.
- By simplifying, the limit calculation will eventually give us the result as \( f'(x) = -\frac{2}{x^2} \).
- This represents how the function changes in value instantaneously at any point, \( x \), not at \( x = 0 \) due to undefined nature.
Slope of Tangent Line
The slope of a tangent line to a function at a particular point indicates the rate at which the y-value of the function is changing with respect to x. It is the value of the derivative at that point. For the function \( f(x) = \frac{2}{x} \), the derivative is given by:\[ f'(x) = -\frac{2}{x^2} \]Calculating specific slopes:
- At \( x = -2 \), \( f'(-2) = -\frac{1}{2} \). This negative slope suggests that the graph is decreasing at this point, and the tangent line dips downwards to the right.
- At \( x = 1 \), \( f'(1) = -2 \). Again, a negative slope indicates a descending graph, with even greater steepness than at \( x = -2 \).
- Notably, \( f'(0) \) is undefined because the original function does not exist at \( x = 0 \).
Other exercises in this chapter
Problem 16
Find \(\frac{d y}{d x}\). $$ y=8 \sqrt{x} $$
View solution Problem 16
Find \(f^{\prime \prime}(x)\) $$ f(x)=x^{1 / 3} $$
View solution Problem 16
Differentiate each function $$ y=\frac{7 x^{3}}{(4-9 x)^{5}} $$
View solution Problem 16
Differentiate two ways: first, by using the Quotient Rule; then, by dividing the expressions before differentiating. Compare your results as a check. Use a grap
View solution