Problem 25
Question
In Exercises \(25-28 :\) a. Find \(f^{-1}(x) .\) b. Graph \(f\) and \(f^{-1}\) together. c. Evaluate \(d f / d x\) at \(x=a\) and \(d f^{-1} / d x\) at \(x=f(a)\) to show that at these points \(d f^{-1} / d x=1 /(d f / d x) .\) $$ f(x)=2 x+3, \quad a=-1 $$
Step-by-Step Solution
Verified Answer
a. \(f^{-1}(x) = \frac{x - 3}{2}\);
b. Graph both as reflections over \(y = x\);
c. \(df/dx = 2, df^{-1}/dx = \frac{1}{2}\) confirms the reciprocal relationship.
1Step 1: Find the Inverse Function
To find the inverse function, \(f^{-1}(x)\), start by setting \(y = f(x) = 2x + 3\). Replace \(f(x)\) with \(y\). Then solve for \(x\) in terms of \(y\): \[ y = 2x + 3 \] Subtract 3 from both sides: \[ y - 3 = 2x \] Divide both sides by 2: \[ x = \frac{y - 3}{2} \] Thus, the inverse function is \(f^{-1}(x) = \frac{x - 3}{2}\).
2Step 2: Graph the Functions
Graph the original function \(f(x) = 2x + 3\) and its inverse \(f^{-1}(x) = \frac{x - 3}{2}\) on the same set of axes. Notice that \(f(x)\) is a straight line with a slope of 2 and y-intercept of 3. The inverse \(f^{-1}(x)\) is also a straight line, but with a slope of \(\frac{1}{2}\) and y-intercept of -1. The lines are reflections of each other over the line \(y = x\).
3Step 3: Differentiate f(x) and f^{-1}(x)
Find \(\frac{df}{dx}\) for \(f(x) = 2x + 3\): \[ \frac{df}{dx} = 2 \] Now find \(\frac{d f^{-1}}{dx}\) for \(f^{-1}(x) = \frac{x - 3}{2}\): \[ \frac{d f^{-1}}{dx} = \frac{1}{2} \]
4Step 4: Evaluate at Specific Points
Evaluate \(\frac{df}{dx}\) at \(x = a = -1\):\[ \frac{df}{dx}\bigg|_{x = -1} = 2 \]Evaluate \(\frac{d f^{-1}}{dx}\) at \(x = f(a) = f(-1) = 1\):\[ \frac{d f^{-1}}{dx}\bigg|_{x = 1} = \frac{1}{2} \]
5Step 5: Confirm the Relationship
Observe the relationship \(\frac{d f^{-1}}{dx} = \frac{1}{\left( \frac{df}{dx} \right)}\) at the specified points.For \(x = a = -1\), \(\frac{df}{dx} = 2\) and at \(x = f(a) = 1\), \(\frac{d f^{-1}}{dx} = \frac{1}{2} = \frac{1}{2}\). This confirms that \(\frac{d f^{-1}}{dx}\) is indeed the reciprocal of \(\frac{df}{dx}\) at these points.
Key Concepts
DifferentiationGraphing FunctionsReciprocal Derivatives
Differentiation
Differentiation is a core concept in calculus that deals with finding the rate at which a quantity changes. In our exercise, we need to differentiate the function \( f(x) = 2x + 3 \). This process is quite straightforward for linear functions like this one. The derivative, \( \frac{df}{dx} \), measures the slope of the tangent line to the curve of the function at any point. For \( f(x) = 2x + 3 \), which is a straight line, the slope is constant. Differentiating, we get \( \frac{df}{dx} = 2 \). This means for every unit increase in \( x \), \( f(x) \) increases by 2.
Next, consider the inverse function, \( f^{-1}(x) = \frac{x - 3}{2} \). To find its derivative, follow a similar process. The result \( \frac{d f^{-1}}{dx} = \frac{1}{2} \) tells us that the inverse function increases by 0.5 for every unit increase in \( x \). This shows how differentiation gives insight into the behavior and change of functions.
Next, consider the inverse function, \( f^{-1}(x) = \frac{x - 3}{2} \). To find its derivative, follow a similar process. The result \( \frac{d f^{-1}}{dx} = \frac{1}{2} \) tells us that the inverse function increases by 0.5 for every unit increase in \( x \). This shows how differentiation gives insight into the behavior and change of functions.
Graphing Functions
Graphing functions is an essential skill in visualizing mathematical concepts, especially when comparing functions and their inverses. Let's graph \( f(x) = 2x + 3 \) along with its inverse \( f^{-1}(x) = \frac{x - 3}{2} \).
The original function \( f(x) \) is a line with a slope of 2 and a y-intercept of 3. This means it rises 2 units vertically for every unit it moves horizontally. On a graph, you would start at the point (0, 3) and draw upwards at a steep angle.
On the other hand, the inverse function \( f^{-1}(x) \) has a slope of \( \frac{1}{2} \), making it less steep. Its y-intercept is -1, meaning you start at the point (0, -1) and rise slowly. Both lines would intersect the line \( y = x \) at the point (1, 1), showcasing that they are mirror images of each other across this line. Graphing these functions concretely illustrates their inverse nature.
The original function \( f(x) \) is a line with a slope of 2 and a y-intercept of 3. This means it rises 2 units vertically for every unit it moves horizontally. On a graph, you would start at the point (0, 3) and draw upwards at a steep angle.
On the other hand, the inverse function \( f^{-1}(x) \) has a slope of \( \frac{1}{2} \), making it less steep. Its y-intercept is -1, meaning you start at the point (0, -1) and rise slowly. Both lines would intersect the line \( y = x \) at the point (1, 1), showcasing that they are mirror images of each other across this line. Graphing these functions concretely illustrates their inverse nature.
Reciprocal Derivatives
In calculus, the concept of reciprocal derivatives often arises when dealing with inverse functions. Understanding this relationship verifies that the inverse functions' slopes are mathematical reciprocals.
In our exercise, we evaluated \( \frac{df}{dx} \) at \( x = -1 \) and found it to be 2. This tells us how sharp the original function's slope is at that point.
Simultaneously, we evaluated \( \frac{d f^{-1}}{dx} \) at \( x = f(-1) = 1 \) and found the result to be \( \frac{1}{2} \). It's crucial to observe how these results confirm the reciprocal relationship: \( \frac{df}{dx} = 2 \) and \( \frac{1}{2} \) are reciprocals.
This relationship indicates a fundamental property of inverses, reflecting both their algebraic and geometric properties. By understanding reciprocal derivatives, students get a deeper insight into how inverse functions refine the graphical reflection over the line \( y = x \).
In our exercise, we evaluated \( \frac{df}{dx} \) at \( x = -1 \) and found it to be 2. This tells us how sharp the original function's slope is at that point.
Simultaneously, we evaluated \( \frac{d f^{-1}}{dx} \) at \( x = f(-1) = 1 \) and found the result to be \( \frac{1}{2} \). It's crucial to observe how these results confirm the reciprocal relationship: \( \frac{df}{dx} = 2 \) and \( \frac{1}{2} \) are reciprocals.
This relationship indicates a fundamental property of inverses, reflecting both their algebraic and geometric properties. By understanding reciprocal derivatives, students get a deeper insight into how inverse functions refine the graphical reflection over the line \( y = x \).
Other exercises in this chapter
Problem 25
In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\theta(\sin (\ln \theta)+\cos (\ln \theta)) $$
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Find the derivative of \(y\) with respect to the given independent variable. \(y=\log _{4} x+\log _{4} x^{2}\)
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In Exercises \(25-36,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\cosh ^{-1} 2 \sqrt{x+1} $$
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In Exercises \(17-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\ln \left(3 \theta e^{-\theta}\right) $$
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