Problem 12
Question
For Problems 7 through 13, find \(f^{\prime}(x), f^{\prime}(0), f^{\prime}(2)\), and \(f^{\prime}(-1) .\) $$ f(x)=\frac{1}{x} $$
Step-by-Step Solution
Verified Answer
The derivative of the function is \(f'(x) = -1/x^2\). \(f'(0)\) is undefined, \(f'(2)\) is -1/4 and \(f'(-1)\) is -1.
1Step 1: Computing the derivative
To differentiate the function \(f(x) = 1/x\), the rule associated with the difference of the power of x is used, specifically, \(d/dx[x^n] = n*x^{(n-1)}\). Here, n = -1. Therefore, the derivative \(f'(x)\) can be expressed as: \(f'(x) = -1 * x^{-2}\) which simplifies to \(f'(x) = -1/x^2\).
2Step 2: Evaluating the derivative at x = 0
The function \(f'(x)\) is undefined at x = 0 because division by 0 is undefined in mathematics. So, \(f'(0)\) is undefined.
3Step 3: Evaluating the derivative at x = 2
Substitute x = 2 into the derivative of the function: \(f'(2) = -1/(2^2)\) which simplifies to \(f'(2) = -1/4\).
4Step 4: Evaluating the derivative at x = -1
Substitute x = -1 into the derivative of the function: \(f'(-1) = -1/((-1)^2)\), which simplifies to \(f'(-1) = -1\).
Key Concepts
DerivativePower RuleFunction Evaluation
Derivative
In calculus, the derivative of a function measures how the function's output changes as its input changes. Imagine you're driving a car; the derivative is like the speedometer telling you how fast you're going at every moment. That's its significance in a broader context. When you take the derivative of a function, you often want to understand how the function behaves and predict its future values.
To differentiate a function means to find this rate of change. For our function, \(f(x) = \frac{1}{x}\), we found the derivative to be \(f'(x) = -\frac{1}{x^2}\). This means that as x changes, the rate at which \(f(x)\) changes is given by \(-\frac{1}{x^2}\). This gives us insight into how drastic or mild these changes are, depending on the value of x.
To differentiate a function means to find this rate of change. For our function, \(f(x) = \frac{1}{x}\), we found the derivative to be \(f'(x) = -\frac{1}{x^2}\). This means that as x changes, the rate at which \(f(x)\) changes is given by \(-\frac{1}{x^2}\). This gives us insight into how drastic or mild these changes are, depending on the value of x.
- The derivative tells us about the slope of the tangent line to the graph of \(f(x)\) at any point \(x\).
- If \(f'(x)\) is positive, \(f(x)\) is increasing; if it's negative, \(f(x)\) is decreasing.
- The magnitude of \(f'(x)\) tells us how steep the curve is at a particular point.
Power Rule
The power rule is a fundamental tool for taking derivatives that simplifies the process greatly. It is essential for students to understand this rule, as it applies in many situations. The power rule, which states \(\frac{d}{dx}[x^n] = n \cdot x^{n-1}\), makes finding the derivative of polynomial functions straightforward.
For the function \(f(x) = \frac{1}{x}\), we recognize this as being \(x^{-1}\), which fits our power rule perfectly with \(n = -1\). By applying the power rule:
For the function \(f(x) = \frac{1}{x}\), we recognize this as being \(x^{-1}\), which fits our power rule perfectly with \(n = -1\). By applying the power rule:
- Differentiate \(x^{-1}\) to get \(-1 \cdot x^{-2}\).
- The derivative simplifies to \(-\frac{1}{x^2}\), as seen in the previous section.
Function Evaluation
Once you have the derivative, evaluating it at certain points can yield critical information about the original function and its behavior. Let's break down how to do this with examples from the solution:
- For \(f'(0)\), we would encounter a division by zero in \(-\frac{1}{x^2}\), meaning \(f'(0)\) is undefined. It indicates a point where the function might have a discontinuity or an asymptote.
- Evaluating \(f'(2)\), you substitute 2 for \(x\) in \(-\frac{1}{x^2}\). This gives \(f'(2) = -\frac{1}{4}\). This indicates that at \(x = 2\), the function is decreasing with a slope of \(-\frac{1}{4}\).
- For \(f'(-1)\), substitute \(-1\) for \(x\) to get \(f'(-1) = -1\). This means at \(x = -1\), the function is decreasing more steeply with a slope of \(-1\).
Other exercises in this chapter
Problem 11
For Problems 7 through 13, find \(f^{\prime}(x), f^{\prime}(0), f^{\prime}(2)\), and \(f^{\prime}(-1) .\) $$ f(x)=x^{2} $$
View solution Problem 12
In Problems 11 and 12, estimate the slope of the tangent line at point \(P\) on the graph of \(f(x)\) by choosing a nearby point \(Q\) on the graph of \(f, Q=(a
View solution Problem 13
For Problems 7 through 13, find \(f^{\prime}(x), f^{\prime}(0), f^{\prime}(2)\), and \(f^{\prime}(-1) .\) $$ f(x)=\frac{x+\pi}{2} $$
View solution Problem 14
If we have a formula for \(f\) we can get quite good numerical estimates of the slope of the tangent line to \(f\) at a particular point. In this exercise we wi
View solution