Chapter 3: Problem 136
For the following exercises, a. evaluate \(f^{\prime}(a),\) and b. graph the function \(f(x)\) and the tangent line at \(x=a\) $$[\mathrm{T}] f(x)=\frac{1}{x}-x^{2 / 3}, a=-1$$
Short Answer
Expert verified
\( f'(a) = -\frac{5}{3} \); graph \( f(x) = \frac{1}{x} - x^{2/3} \) and the line \( y = -\frac{5}{3}x - \frac{11}{3} \).
Step by step solution
01
Find the derivative f'(x)
The function is given by \( f(x) = \frac{1}{x} - x^{2/3} \). First, find the derivative \( f'(x) \) using the power rule and the derivative of \( \frac{1}{x} \), which is \( -\frac{1}{x^2} \). Thus, the derivative is:\[ f'(x) = -\frac{1}{x^2} - \frac{2}{3}x^{-1/3}. \]
02
Evaluate the derivative at x = a
We need to evaluate \( f'(x) \) at \( a = -1 \). Substitute \( x = -1 \) into the derivative:\[ f'(-1) = -\frac{1}{(-1)^2} - \frac{2}{3}(-1)^{-1/3} = -1 - \frac{2}{3}(-1)^2 = -1 - \frac{2}{3} = -\frac{5}{3}. \]
03
Calculate f(a) for the tangent line
Substitute \( a = -1 \) into the original function to find \( f(-1) \):\[ f(-1) = \frac{1}{-1} - (-1)^{2/3} = -1 - 1 = -2. \]
04
Equation of the tangent line
The equation of a line is \( y = mx + b \). Here, the slope \( m = f'(-1) = -\frac{5}{3} \) and the line passes through the point \( (-1, -2) \). Use the point-slope form: \[ y + 2 = -\frac{5}{3}(x + 1). \] Simplify to get the equation in slope-intercept form:\[ y = -\frac{5}{3}x - \frac{11}{3}. \]
05
Graph f(x) and the tangent line
Graph the original function \( f(x) = \frac{1}{x} - x^{2/3} \) and the tangent line \( y = -\frac{5}{3}x - \frac{11}{3} \). The function is not defined at \( x = 0 \), so split the graph into two parts. The tangent line touches the graph of the function at \( x = -1 \) with the value \( f(-1) = -2 \). Note that the line \( y = -\frac{5}{3}x - \frac{11}{3} \) should touch \( f(x) \) exactly at this point.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
When studying calculus, the concept of a derivative is fundamental. A derivative represents the rate at which a function changes at any point. It's like the speedometer in your car, showing how fast something changes. In formal terms, if you have a function \( f(x) \), its derivative \( f'(x) \) tells you the slope of the tangent line at any \( x \) value.
For example, the function given here is \( f(x) = \frac{1}{x} - x^{2/3} \). To find its derivative \( f'(x) \), you must understand how each part of \( f(x) \) behaves when differentiated. For \( \frac{1}{x} \), the derivative is \( -\frac{1}{x^2} \). The term \( x^{2/3} \) involves using the power rule, which we will touch on later. Together, they form the overall derivative:
For example, the function given here is \( f(x) = \frac{1}{x} - x^{2/3} \). To find its derivative \( f'(x) \), you must understand how each part of \( f(x) \) behaves when differentiated. For \( \frac{1}{x} \), the derivative is \( -\frac{1}{x^2} \). The term \( x^{2/3} \) involves using the power rule, which we will touch on later. Together, they form the overall derivative:
- \( -\frac{1}{x^2} \) from \( \frac{1}{x} \)
- \( -\frac{2}{3}x^{-1/3} \) from \( -x^{2/3} \)
Tangent Line
The concept of a tangent line is closely linked to derivatives. Imagine a straight line that just "touches" a curve at a single, specific point, without crossing it. This is the tangent line, and it shares the same slope as the curve at the point of contact.
To find the equation of a tangent line to a function at a specific \( x \) value, we use the derivative. It's crucial because the derivative gives us the slope of the function at that point. In our exercise, we calculated the slope, which is \( f'(-1) = -\frac{5}{3} \). Using this in the point-slope form equation of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is our slope and \( (x_1, y_1) \) is the point of tangency, \( (-1, -2) \), we have:
To find the equation of a tangent line to a function at a specific \( x \) value, we use the derivative. It's crucial because the derivative gives us the slope of the function at that point. In our exercise, we calculated the slope, which is \( f'(-1) = -\frac{5}{3} \). Using this in the point-slope form equation of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is our slope and \( (x_1, y_1) \) is the point of tangency, \( (-1, -2) \), we have:
- \( y + 2 = -\frac{5}{3}(x + 1) \)
- Simplified: \( y = -\frac{5}{3}x - \frac{11}{3} \)
Graphing Functions
Graphing functions is a visual way to understand the behavior of functions over different intervals. It allows us to see patterns, trends, and points where the function behaves in a particular way. Let's consider our function \( f(x) = \frac{1}{x} - x^{2/3} \).
When graphing, look out for values where the function is not defined. For us, \( f(x) \) is undefined at \( x = 0 \) because \( \frac{1}{x} \) becomes problematic. This means the graph will have two separate parts: one for negative values and one for positive values.
Also, note the interaction with the tangent line \( y = -\frac{5}{3}x - \frac{11}{3} \). This line meets the graph at the point \( x = -1 \), \( y = -2 \). By plotting both the function and the tangent line together, you observe:
When graphing, look out for values where the function is not defined. For us, \( f(x) \) is undefined at \( x = 0 \) because \( \frac{1}{x} \) becomes problematic. This means the graph will have two separate parts: one for negative values and one for positive values.
Also, note the interaction with the tangent line \( y = -\frac{5}{3}x - \frac{11}{3} \). This line meets the graph at the point \( x = -1 \), \( y = -2 \). By plotting both the function and the tangent line together, you observe:
- The curve of the function touches the tangent line at this specific point.
- Elsewhere, the two separate lines never meet, highlighting how the derivative's information localizes the tangent exactly to its point of contact.
Power Rule
The power rule is a straightforward yet powerful tool for finding derivatives. It's especially useful when dealing with powers of \( x \) in functions. The rule states: if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \). This simplifies derivative calculations enormously.
Using the power rule, we can find the derivative of \( x^{2/3} \). Applying \( f(x) = x^{2/3} \), the rule gives: \( f'(x) = \frac{2}{3}x^{-1/3} \). This transforms challenging algebra into a manageable form.
In our exercise, the power rule helped break down \( f(x) = \frac{1}{x} - x^{2/3} \):
Using the power rule, we can find the derivative of \( x^{2/3} \). Applying \( f(x) = x^{2/3} \), the rule gives: \( f'(x) = \frac{2}{3}x^{-1/3} \). This transforms challenging algebra into a manageable form.
In our exercise, the power rule helped break down \( f(x) = \frac{1}{x} - x^{2/3} \):
- The term \( x^{2/3} \) yielded its derivative \( -\frac{2}{3}x^{-1/3} \).
- Coupled with \( \frac{1}{x} \)'s derivative of \( -\frac{1}{x^2} \), we derive \( f'(x) = -\frac{1}{x^2} - \frac{2}{3}x^{-1/3} \).