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Chapter 23: Problem 36

In Exercises \(35-40,\) solve the given problems. Find the point(s) on the curve of \(y=1 /(x+1)\) for which the slope of the tangent line is -1

Short Answer

Expert verified
The points are \((0, 1)\) and \((-2, -1)\).

Step by step solution

01

Identify the Derivative

To find the point where the slope of the tangent line is -1, begin by finding the derivative of the function. Given the function \( y = \frac{1}{x+1} \), use the power rule and the chain rule to differentiate it. The derivative \( y' \) is the slope of the tangent at any point \( x \).\[ y' = \frac{d}{dx} \left( (x+1)^{-1} \right) \]This simplifies to: \[ y' = - (x+1)^{-2} = - \frac{1}{(x+1)^2}\]
02

Set Derivative Equal to -1

Set the derivative equal to -1, since we are searching for when this slope occurs:\[ - \frac{1}{(x+1)^2} = -1\]Solving this equation will give us the \( x \)-value where the slope of the tangent line is -1.
03

Solve the Equation

Solve the equation \(- \frac{1}{(x+1)^2} = -1\):Eliminate the negative sign and solve for \(x\):\[ \frac{1}{(x+1)^2} = 1 \]Taking the reciprocal, we have:\[ (x+1)^2 = 1 \]Find the square root of both sides:\[ x+1 = \pm 1 \]
04

Find the x-values

Solve for \(x\) from \(x+1 = 1\) and \(x+1 = -1\):1. \(x+1 = 1\) gives \(x = 0\)2. \(x+1 = -1\) gives \(x = -2\)Thus, the \(x\)-values are \(x = 0\) and \(x = -2\).
05

Calculate the y-values

Using the original function \( y = \frac{1}{x+1} \), substitute \( x = 0 \) and \( x = -2 \) to find the corresponding \( y \)-values:1. At \( x = 0 \), \( y = \frac{1}{0+1} = 1 \)2. At \( x = -2 \), \( y = \frac{1}{-2+1} = -1 \)The points where the slope of the tangent is -1 are \((0, 1)\) and \((-2, -1)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
A derivative represents the rate at which a function changes at any given point. It's a fundamental concept in calculus and is essential for determining the slope of a tangent line to a curve. To find the derivative, you're looking for a new function that describes the instantaneous rate of change of the original function.
For the function \( y = \frac{1}{x+1} \):
  • First, rewrite the function to a power that is easier to differentiate: \((x+1)^{-1}\).
  • Utilize the power rule, which involves bringing down the exponent as a multiplier and reducing the exponent by one. For a function \((x+1)^{-n}\), its derivative is \(-n(x+1)^{-(n+1)}\).
  • This gives us the derivative \( y' = - (x+1)^{-2} = - \frac{1}{(x+1)^2}\).
Understanding how to find a derivative is key in determining tangents and slopes along a function's curve.
Tangent Line
A tangent line touches a curve at a single point and its slope equals the derivative of the function at that point. Tangent lines are incredibly useful as they provide an approximation of the curve at a specific point.
In the context of the problem, the tangent line's slope is specified to be \(-1\). This means:
  • You need to find where the derivative \( y' = - \frac{1}{(x+1)^2} \) equals \(-1\).
  • Set the derivative equal to \(-1\) and solve \(- \frac{1}{(x+1)^2} = -1\) to determine the precise points on the curve.
This provides key insights into how the curve behaves at specific locations.
Function Slope
The slope of a function at a given point informs us about the steepness and direction in which the function moves. In other words, it's a measure of how much the function \( y \) changes with respect to changes in \( x \).
  • In this problem's context, a slope of \(-1\) indicates a decrement: as \( x \) increases by 1 unit, \( y \) decreases by 1 unit.
  • Understanding the slope helps identify whether a function is increasing or decreasing at a point.
Knowing the slope is vital when analyzing a function's behavior and predicting its curve鈥檚 direction and shape.
Chain Rule
The chain rule is a method for differentiating composite functions. When dealing with such functions, the chain rule is essential because it allows you to take derivatives of the inner and outer functions separately.
  • For example, the function \( y = \frac{1}{x+1} \) can be seen as the composition of two functions: the outer function \( u^{-1} \), and the inner function \( u = x+1 \).
  • Applying the chain rule involves differentiating the outer function with respect to the inner function, and then multiplying it by the derivative of the inner function.
  • Through the chain rule, we end up with the derivative \( y' = - \frac{1}{(x+1)^2} \).
The chain rule is crucial for dealing with more complex functions and enables accurate derivative computation for all involved function parts.

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