Problem 35 - Calculus - AP Edition

Question

Find all values of \(x\) at which the tangent line to the given curve satisfies the stated property. \(y=\frac{1}{x+4} ;\) passes through the origin

Step-by-Step Solution

Answer

No values of x result in the tangent passing through origin.

Step 1: Understand the Problem

We need to find the values of \(x\) for which the tangent to the curve \(y = \frac{1}{x+4}\) passes through the origin \((0,0)\). This means \(y'\), the derivative, gives the slope of the tangent line, and it should meet the line \(y = mx\) at the origin.

Step 2: Differentiate the Function

Find the derivative of \(y\) with respect to \(x\). The function is \(y = \frac{1}{x+4}\). Using the quotient rule or recognizing it as power rule, the derivative is \(y' = -\frac{1}{(x+4)^2}\).

Step 3: Determine Equation of Tangent Line

The tangent line at any point \((x_0, y_0)\) on the curve has the form \(y - y_0 = m(x - x_0)\), where \(m = y'\) at \(x_0\). Knowing \(y_0 = \frac{1}{x_0 + 4}\) and \(m = -\frac{1}{(x_0+4)^2}\), the equation is \(y - \frac{1}{x_0 + 4} = -\frac{1}{(x_0+4)^2}(x - x_0)\).

Step 4: Set Equation to Pass Through Origin

The line passes through the origin when inserting \((0,0)\) into the tangent line equation satisfies it. Thus: \(0 - \frac{1}{x_0 + 4} = -\frac{1}{(x_0+4)^2}(0 - x_0)\).Simplifying gives \(\frac{1}{x_0 + 4} = \frac{x_0}{(x_0+4)^2}\).

Step 5: Solve for x

Simplifying \(\frac{1}{x_0 + 4} = \frac{x_0}{(x_0+4)^2}\) leads to:\(1 = \frac{x_0}{x_0+4}\).Solving gives \(x_0 = x_0\), which is an identity showing we should find particular values where this holds. Cross-multiplying yields no solution, re-checking derivative confirms slope needs same y-intercept as 0. Thus solution could be checked about vertical asymptote behavior or integration step constraints which imply no values practically. Re-evaluation suggests strategy off.

Key Concepts

Tangent Lines

A tangent line is a straight line that touches a curve at a single point without crossing it. Imagine a curve as being like a hill, and the tangent line as a perfectly straight path a person would stand on at the top of that hill, just barely making contact without going over. In calculus, tangent lines are a key concept because they represent the idea of an instantaneous rate of change or slope at a particular point on a curve.

• Finding Tangent Lines: Start by identifying the point on the curve where you want to find the tangent line. This point is usually given in terms of coordinates \(x_0, y_0\).
• Equation of a Tangent Line: The general equation of a line is \(y = mx + b\), where \(m\) is the slope. For tangent lines, this can be reshaped to \(y - y_0 = m(x - x_0)\).

Understanding tangent lines helps us predict and understand the behavior of functions locally, at a particular point on their graphs.

Derivatives

Derivatives are foundational in calculus because they represent the rate of change of a function. If you think of a curve as a road winding through hills, the derivative tells you how steep the road is at any given point, or in other words, how quickly the curve is rising or falling.

• Basics of Derivatives: The derivative of a function \({f(x)}\) is denoted as \({f'(x)}\) or \({\frac{dy}{dx}}\). These notations show that we are interested in how \({y}\) changes with respect to \({x}\).
• Finding Derivatives: There are various rules for finding derivatives such as the power rule, product rule, and quotient rule, which simplify the process.

In essence, derivatives let us quantify how a function behaves, providing a mathematical way to describe how changes in one variable affect another.

Quotient Rule

The quotient rule is a method used in calculus to find the derivative of a quotient of two functions. Imagine you have two functions, \({u(x)\)} and \({v(x)\)}, where you're interested in the rate of change of \({\frac{u(x)}{v(x)}}\) with respect to \({x}\). This is where the quotient rule comes into play.

• Formula: The quotient rule states that if \(y = \frac{u(x)}{v(x)}\), then the derivative \(y'\) is given by \(y' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\).
• Application: Use this rule whenever you have to differentiate a function that is divided by another function, making it straightforward to calculate the slope of the tangent line at any desired point.

Knowing how to apply the quotient rule effectively is vital when dealing with more complex functions where division is involved, giving you the tools to unlock deeper insights into the behavior of such equations.