Problem 10
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: Understand the Relationship between a Function and Its Inverse
For a function \( g(x) \) and its inverse \( g^{-1}(x) \), the slopes of the tangent lines at corresponding points are multiplicative inverses of each other. This means that if the slope of \( g(x) \) is \( m \) at a certain point, then the slope of \( g^{-1}(x) \) is \( \frac{1}{m} \) at the corresponding point on the inverse.
2Step 2: Identify Given Information
The function \( g(x) \) passes through the origin, which means \( g(0) = 0 \). The slope of \( g(x) \) at the origin is given as 2.
3Step 3: Apply the Slope Relationship
Since the slope of \( g(x) \) at the origin \((0, 0)\) is 2, the slope of the inverse function \( g^{-1}(x) \) at the corresponding point \( (0, 0) \) is \( \frac{1}{2} \). This is because the slopes of a function and its inverse at corresponding points are reciprocals of each other.
Key Concepts
Differentiable FunctionSlope of a Tangent LineFunction Inverses
Differentiable Function
A differentiable function is one that has a derivative at every point in its domain. This means it's a smooth curve, without any sharp corners or breaks. The derivative gives us the rate of change of the function at any specific point. For example, with our function \( y = g(x) \), it's defined as differentiable, which indicates that we can determine its slope anywhere on its graph.
When a function is differentiable, it allows us to talk about how the function is changing at any particular point. This is crucial because it means the function behaves predictably. If you imagine drawing a tangent line at any point on the graph of a differentiable function, that line would only touch the graph at one point and follow the exact direction it's moving locally.
The concept of differentiability is significant for finding slopes, which we'll discuss next, and also for understanding how the function's inverse will behave.
When a function is differentiable, it allows us to talk about how the function is changing at any particular point. This is crucial because it means the function behaves predictably. If you imagine drawing a tangent line at any point on the graph of a differentiable function, that line would only touch the graph at one point and follow the exact direction it's moving locally.
The concept of differentiability is significant for finding slopes, which we'll discuss next, and also for understanding how the function's inverse will behave.
Slope of a Tangent Line
The slope of a tangent line to a function at a given point tells us the rate at which the function's value is changing at that point. In mathematical terms, this slope is represented by the derivative. For our function \( y = g(x) \), the derivative, denoted usually as \( g'(x) \), gives us this slope.
At any specific point, the slope of the tangent line is the same as the function's derivative at that point. If the slope is positive, like 2 in our original example, it tells us the function is increasing as you move along the x-axis. A negative slope would indicate the function is decreasing.
Knowing the slope of a tangent line is crucial for understanding how rapid the changes are happening in the function's graph. For any function and its inverse, understanding the slope is essential to determining how the graphs of those two functions relate to each other.
At any specific point, the slope of the tangent line is the same as the function's derivative at that point. If the slope is positive, like 2 in our original example, it tells us the function is increasing as you move along the x-axis. A negative slope would indicate the function is decreasing.
Knowing the slope of a tangent line is crucial for understanding how rapid the changes are happening in the function's graph. For any function and its inverse, understanding the slope is essential to determining how the graphs of those two functions relate to each other.
- Positive slope: Indicates an increasing function.
- Negative slope: Indicates a decreasing function.
- Zero slope: Represents a horizontal tangent, where no vertical change occurs.
Function Inverses
Function inverses represent a sort of mirror image of the original function across the line \( y = x \). If you take a function \( g(x) \) and its inverse \( g^{-1}(x) \), the output of the original function becomes the input for the inverse, and vice versa.
Importantly, the slope of the original function at any point is the reciprocal of the slope of its inverse at the corresponding point. For example, if \( g(x) \) has a slope of 2 at a point, \( g^{-1}(x) \) would have a slope of \( \frac{1}{2} \) at the corresponding point. This relationship is pivotal because it helps us understand how the two functions transition from one to the other.
In practice, this means:
Importantly, the slope of the original function at any point is the reciprocal of the slope of its inverse at the corresponding point. For example, if \( g(x) \) has a slope of 2 at a point, \( g^{-1}(x) \) would have a slope of \( \frac{1}{2} \) at the corresponding point. This relationship is pivotal because it helps us understand how the two functions transition from one to the other.
In practice, this means:
- The steeper the slope of the function, the flatter the slope of its inverse at corresponding points.
- If a function has a slope of zero at a point, its inverse does not have a defined slope at the corresponding point, implying vertical tangent lines.
Other exercises in this chapter
Problem 10
If \(r+s^{2}+v^{3}=12, d r / d t=4,\) and \(d s / d t=-3,\) find \(d v / d t\) when \(r=3\) and \(s=1\).
View solution Problem 10
Find a linearization at a suitably chosen integer near \(a\) at which the given function and its derivative are easy to evaluate. $$f(x)=1+x, \quad a=8.1$$
View solution Problem 10
In Exercises \(9-22,\) write the function in the form \(y=f(u)\) and \(u=g(x) .\) Then find \(d y / d x\) as a function of \(x\). $$y=(4-3 x)^{9}$$
View solution Problem 10
Use implicit differentiation to find \(d y / d x\). $$x y=\cot (x y)$$
View solution