Problem 34
Question
Suppose that the differentiable function \(y=g(x)\) has an inverse and that the graph of \(g\) passes through the origin with slope \(2 .\) Find the slope of the graph of \(g^{-1}\) at the origin.
Step-by-Step Solution
Verified Answer
The slope of \(g^{-1}\) at the origin is \(\frac{1}{2}\).
1Step 1: Understanding the Inverse Function Relationship
If a function \( y = g(x) \) has an inverse, then it can be written as \( x = g^{-1}(y) \). The relationship between the slopes of these functions at any points \((a, b)\) and \((b, a)\) respectively is given by the formula:\[ (g^{-1})'(b) = \frac{1}{g'(a)}\]where \(g(a) = b\).
2Step 2: Finding Given Information
From the problem, we know that the function \( g(x)\) passes through the origin, which means \( g(0) = 0 \), and the slope of \( g \) at the origin is \( g'(0) = 2 \).
3Step 3: Applying the Inverse Derivative Formula
Since the graph of \( g \) at the origin has slope 2, and considering the formula for the derivative of an inverse function, we can substitute \(a = 0\), \(b = 0\), and \(g'(0) = 2 \) into:\[ (g^{-1})'(0) = \frac{1}{g'(0)} = \frac{1}{2}\]
4Step 4: Conclusion
The slope of the graph of the inverse function \( g^{-1} \) at the origin is \( \frac{1}{2} \).
Key Concepts
Differentiable FunctionSlope of a FunctionDerivative Formula
Differentiable Function
A differentiable function is one that has a derivative at every point in its domain. This means you can calculate the rate at which the function's value changes with respect to changes in its input. Differentiability is a key feature that ensures smoothness in the graph of the function, meaning it has no sharp corners or cusps.
If a function is differentiable at a point, it is also continuous at that point. However, continuity does not imply differentiability. Differentiable functions are important in calculus because they allow us to use derivatives to analyze the behavior of functions deeply.
Whenever working with an inverse function like in the exercise provided, differentiability helps in finding the slope of an inverse graph through the derivative of the original function.
If a function is differentiable at a point, it is also continuous at that point. However, continuity does not imply differentiability. Differentiable functions are important in calculus because they allow us to use derivatives to analyze the behavior of functions deeply.
Whenever working with an inverse function like in the exercise provided, differentiability helps in finding the slope of an inverse graph through the derivative of the original function.
Slope of a Function
The slope of a function at a given point is the same as the derivative of the function at that point. It gives information about the steepness or inclination of the curve at that specific point. The slope is mathematically calculated as the ratio of the change in the function's output to the change in its input, as you move infinitesimally along the graph.
- A positive slope indicates that the function is increasing at that point.
- A negative slope means the function is decreasing.
- A slope of zero signifies that the function has a horizontal tangent, possibly at a local maximum or minimum.
Derivative Formula
The derivative formula is a mathematical expression that gives the derivative of a function. The derivative represents the function's rate of change, and in simpler terms, its slope at any given point.
The most fundamental derivative formula is usually derived as follows:
\[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \]
This represents the derivative of the function \( f(x) \) as \( h \) approaches zero. It essentially captures how much the function \( f(x) \) changes when the input \( x \) changes by a tiny amount \( h \).
When dealing with inverse functions, the derivative formula holds the key to understanding the slope of the inverse graph. As explained in this exercise, for differentiable functions, if \( y = g(x) \) has an inverse \( x = g^{-1}(y) \), then the slope of the inverse at a point can be found using the formula:
\[ (g^{-1})'(b) = \frac{1}{g'(a)} \]
Here, \( g(a) = b \), making it crucial to know both the function and its inverse to use this formula effectively.
The most fundamental derivative formula is usually derived as follows:
\[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \]
This represents the derivative of the function \( f(x) \) as \( h \) approaches zero. It essentially captures how much the function \( f(x) \) changes when the input \( x \) changes by a tiny amount \( h \).
When dealing with inverse functions, the derivative formula holds the key to understanding the slope of the inverse graph. As explained in this exercise, for differentiable functions, if \( y = g(x) \) has an inverse \( x = g^{-1}(y) \), then the slope of the inverse at a point can be found using the formula:
\[ (g^{-1})'(b) = \frac{1}{g'(a)} \]
Here, \( g(a) = b \), making it crucial to know both the function and its inverse to use this formula effectively.
Other exercises in this chapter
Problem 34
In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\ln \sqrt{\frac{(x+1)^{5}}{(x+2)^{20}}} $$
View solution Problem 34
Find the derivative of \(y\) with respect to the given independent variable. \(y=\log _{2}\left(\frac{x^{2} e^{2}}{2 \sqrt{x+1}}\right)\)
View solution Problem 35
In Exercises \(25-36,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\sinh ^{-1}(\tan x) $$
View solution Problem 35
In Exercises \(17-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\int_{0}^{\ln x} \sin e^{t} d t $$
View solution