Problem 71 - Calculus Single Variable

Question

The graph of \(f(x)=2 x /(x+1)\) and the point \(P=(2,4)\) which lies above the graph, are shown in Figure \(3 .\) The purpose of this exercise is to find the equation of every tangent line to the graph of \(f\) that passes through \(P\). Copy Figure 3 on your paper. To your sketch, add the tangent lines that pass through \(P\). How many do there appear to be? You will use this predicted number \(N\) as a "reality check" in part \(c\). a. What is the slope of the tangent line to the graph of \(f\) at \((c, f(c)) ?\) (By inspection, the tangents to the graph of \(f\) all slope upward. Before continuing, make sure that your algebraic formula for the slope is consistent with this property.) b. What is the equation of the tangent line to the graph of \(f\) at \((c, f(c))\) c. What equation must \(c\) satisfy if the tangent line at \((c, f(c))\) passes through \(P ?\) (How many solutions does this equation have? Before continuing, make sure that the equation for \(c\) has the same number of solutions as your predicted number \(N .)\) d. Find the equations of all tangent lines to the graph of \(f\) that pass through \(P\).

Step-by-Step Solution

Answer

There are two tangent lines passing through \((2,4)\).

Step 1: Find the derivative of f(x)

To find the tangent line's slope at any point \((c, f(c))\), we first need \(f'(x)\). The function is \(f(x) = \frac{2x}{x+1}\). Use the quotient rule: if \(u(x)=2x\) and \(v(x)=x+1\), then \(f'(x)=\frac{u'v - uv'}{v^2}\). We have \(u'=2\) and \(v'=1\), so \(f'(x)=\frac{2(x+1) - 2x}{(x+1)^2}=\frac{2}{(x+1)^2}\). This indicates the slope of the tangent at any point \((c, f(c))\) is \(f'(c)=\frac{2}{(c+1)^2}\).

Step 2: Write the equation of the tangent line at (c, f(c))

The tangent line at \((c, f(c))\) to \(f(x)\) has a general form: \(y - f(c) = f'(c)(x - c)\). With \(f(c) = \frac{2c}{c+1}\) and \(f'(c) = \frac{2}{(c+1)^2}\), the equation becomes: \[y - \frac{2c}{c+1} = \frac{2}{(c+1)^2}(x - c)\].

Step 3: Find condition for tangent through P=(2, 4)

The tangent passes through \((2, 4)\), substituting into the line equation: \[4 - \frac{2c}{c+1} = \frac{2}{(c+1)^2}(2 - c)\]. Simplifying: \[ \frac{4(c+1) - 2c}{c+1} = \frac{2(2-c)}{(c+1)^2}\]. Solving this will give conditions on \(c\).

Step 4: Solve for c

Simplify and solve the equation: \[ \frac{4c + 4 - 2c}{c+1} = \frac{4 - 2c}{(c+1)^2}\]. -> \[ \frac{2c + 4}{c+1} = \frac{4-2c}{(c+1)^2}\]. Multiply through by \((c+1)^2\) to get \((2c+4)(c+1) = 4 - 2c\), which simplifies to \(2c^2 + 2c + 6 = 0\). Solve for \(c\) using the quadratic formula: \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), yielding two potential \(c\) values.

Step 5: Verify number of solutions and compute tangent equations

We find two solutions indicating two possible tangent lines, as the predicted \(N\) should match the number of tangents we saw intersect \(P\). Substitute these \(c\) values back to find explicit equations for each tangent through \((2, 4)\).

Key Concepts

Quotient Rule

When dealing with functions that are expressed as the quotient of two functions, like our function \( f(x) = \frac{2x}{x+1} \), we use the quotient rule to find the derivative. The quotient rule is a method in calculus used to differentiate functions that are a division of two differentiable functions. It states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their quotient \( \frac{u}{v} \) is given by:\[f'(x) = \frac{u'v - uv'}{v^2}\]- **\( u(x) \)** is the numerator of your function, for \( f(x) \) here, it is \( 2x \).- **\( v(x) \)** is the denominator, which, in this case, is \( x+1 \).- **\( u'(x) \)** and **\( v'(x) \)** are the derivatives of \( u(x) \) and \( v(x) \) respectively.In the problem, we find that \( u' = 2 \) and \( v' = 1 \). Substituting these into the quotient rule provides us:\[f'(x) = \frac{2(x+1) - 2x}{(x+1)^2} = \frac{2}{(x+1)^2}\]Understanding and applying the quotient rule is essential in finding the slope of tangent lines for rational functions.

Derivative Calculation

Derivatives allow us to find the slope of a function at any given point, which is crucial when searching for tangent lines. In the context of this problem, we needed the derivative of \( f(x) = \frac{2x}{x+1} \) to establish the slope of any tangent line intersecting at point \((c, f(c))\). After applying the quotient rule, we found:\[f'(x) = \frac{2}{(x+1)^2}\]This result shows the slope of the function at each point \( (c, f(c)) \) and lets us determine the tangent line. For any function, knowing how to compute the derivative—whether it's via the quotient rule, product rule, chain rule, or simply differentiating—is vital for solving similar problems.

• Derivatives provide rate of change.
• Essential for finding tangent slopes.
• Applied in various mathematical and physical concepts.

Quadratic Equation

Quadratic equations often appear in calculus when solving problems involving curves and tangency. After differentiating and working through the problem, we arrive at the equation:\[2c^2 + 2c + 6 = 0\]This is a quadratic equation in standard form \( ax^2 + bx + c = 0 \), where \( a = 2 \), \( b = 2 \), and \( c = 6 \). When solving, we utilize the quadratic formula:\[c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Quadratic equations can have zero, one, or two solutions depending on the discriminant \( b^2 - 4ac \). Here, solving this quadratic gives us the potential values of \( c \) needed for our tangent equations. Recognizing and solving quadratic equations correctly is essential in calculus for understanding intersections and tangencies.

Graph of a Function

The graph of a function provides visual insight into how the function behaves and how tangent lines intersect with it. Given in the problem, function \( f(x) = \frac{2x}{x+1} \) is a rational function, typically showing hyperbolic behavior. The graph shows the curve's general direction and helps highlight significant points of interest such as the tangent.For this particular function:

• The graph slants upwards, demonstrating a positive slope.
• Being a hyperbola, it has vertical and horizontal asymptotes.
• Tangent lines appear to touch the curve precisely at a single point and can be extended to intersect external points, such as \(P = (2,4)\).

Understanding how to plot and interpret the graph gives students tangible insight into where tangent lines might occur, how many tangents can pass through an external point, and the general behavior of rational functions.