Question

a. Use definition ( 2 ) ( \(p .\) 135 ) to find the slope of the line tangent to the graph of \(f\) at \(P\). b. Determine an equation of the tangent line at \(P\). $$f(x)=\frac{x}{x+1} ; P(-2,2)$$

Short answer

a) The slope of the tangent line to the graph of \(f(x)\) at point P(\(-2,2\)) is -3. b) The equation of the tangent line at point P is \(y = -3x - 4\).

Step-by-step solution

Describe the problem

We need to find the slope of the tangent line to the graph of f(x) at P as well as the equation of the tangent line.

Find derivative of the function

First, we need to find the derivative of the function, \(f'(x)\). The function is: $$f(x)=\frac{x}{x+1}$$ We can rewrite it as a product of two functions, \(u(x)=x\) and \(v(x)=\frac{1}{x+1}\) (where \(v(x)\) is the reciprocal of the sum of \(x\) and 1). Now, we'll use the product rule to find the derivative of \(f(x)\). The product rule states: $$(u \cdot v)' = u' \cdot v + u \cdot v'$$

Compute derivatives of u(x) and v(x)

Now, let's find the derivatives of \(u(x)\) and \(v(x)\). For \(u(x) = x\), the derivative is \(u'(x) = 1\). For \(v(x) = \frac{1}{x+1}\), we need to find \(v'(x)\). Note that, \(v(x) = (x+1)^{-1}\), and since it is a simple power function, the derivative is found by multiplying the exponent with the function and then subtracting one from the exponent: $$v'(x) = -1 (x+1)^{-2}$$

Apply the product rule

Now, apply the product rule to find the derivative of the function \(f'(x)\): $$f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$$ $$f'(x) = 1 \cdot \frac{1}{x+1} + x \cdot (-1)(x+1)^{-2}$$ Simplify: $$f'(x) = \frac{1}{x+1} - \frac{x}{(x+1)^2}$$

Find the slope at the given point P

Now, we can find the slope of the tangent line at P by plugging in the \(x\)-value of point P, which is \(-2\), into the derivative: $$f'(-2) = \frac{1}{(-2)+1} - \frac{(-2)}{(-2+1)^2}$$ $$f'(-2) = -1 - 2 = -3$$ So the slope of the tangent line at P is -3.

Find the equation of the tangent line

Lastly, to find the equation of the tangent line, we can use the point-slope form of the equation: \(y - y_1 = m(x - x_1)\). We know the point P is \((-2,2)\) and the slope \(m = -3\). Plug in these values: $$y - 2 = -3(x - (-2))$$ Simplify: $$y - 2 = -3(x + 2)$$ $$y = -3x - 6 + 2$$ $$y = -3x - 4$$ So the equation of the tangent line at point P is \(y = -3x - 4\).

Key concepts

Derivative

Derivatives are fundamental in calculus, helping us determine how a function changes at any given point. They show the rate at which \(f(x)\) is changing when its input \(x\) changes. For the function \(f(x) = \frac{x}{x+1}\), the derivative \(f'(x)\) tells us the slope of the line tangent to the curve at any point \(x\). This derivative emerges from the limit definition, capturing how small changes in \(x\) affect \(f(x)\).
In our exercise, computing \(f'(x)\) involves differentiating a function expressed as a division, which we can rewrite as a product using two separate functions, \(u(x) = x\) and \(v(x) = \frac{1}{x+1}\). This simplification allows us to utilize further methods like the Product Rule.

Product Rule

The Product Rule is a technique used to find the derivative of the product of two functions. If we have two functions, \(u(x)\) and \(v(x)\), their derivative is given by:

• \( (u \cdot v)' = u' \, v + u \, v' \)

In our example, we apply the Product Rule to \(f(x) = x \cdot \frac{1}{x+1}\). Breaking it down:

• \(u(x) = x\) has a derivative \(u'(x) = 1\).
• \(v(x) = (x+1)^{-1}\), which differentiates to \(v'(x) = -1(x+1)^{-2}\).

By plugging these into the Product Rule, we find \(f'(x)\), which helps us evaluate the slope at any point, an essential step to understanding the tangent line.

Tangent Line Equation

The tangent line to a curve at a given point is a straight line that just touches the curve at that point, matching the curve's slope there. To find its equation, we usually apply the point-slope form: $$ y - y_1 = m(x - x_1) $$ Here, $(x_1, y_1)$ is the point on the curve where the tangent touches, and $m$ is the slope at that point.
For our function $f(x)$ at point $P(-2, 2)$, we have calculated the slope, $m = -3$, using the derivative. By substituting these values into the point-slope form, we work through the algebra to yield $y = -3x - 4$, giving us the tangent line's equation. This line represents the linear approximation of the curve at the given point, $P$.

Slope of Tangent Line

The slope of the tangent line is crucial because it tells us how steep the line is at the point it touches the curve. It's essentially the rate of change of the function at that specific point.
In our exercise, after finding the derivative \(f'(x) = \frac{1}{x+1} - \frac{x}{(x+1)^2}\), we determined the slope at \(P(-2, 2)\) to be \(-3\). We did this by substituting \(x = -2\) into the derivative.
This negative slope tells us that, at point \(P\), the function is decreasing. Understanding this step is key to visualizing how the function behaves around \(P\) and predicting its path beyond the tangent point.