Chapter 3: Problem 26
Equations of tangent lines by definition (2) a. Use definition (2) ( \(p .\) 129) 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
Expert verified
Answer: The equation of the tangent line is \(y = -6x - 10\).
Step by step solution
01
Compute the derivative of the function
We need to find the derivative of \(f(x)=\frac{x}{x+1}\). To do that, first, we will use the definition:
$$ f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} $$
02
Plug \(f(x+h)\) and \(f(x)\) into the definition
Now replace \(f(x+h)\) and \(f(x)\) in the definition with the given function:
$$f'(x) = \lim_{h\rightarrow 0} \frac{\frac{x+h}{x+h+1} - \frac{x}{x+1}}{h}$$
03
Simplify the difference between two fractions
To eliminate the fractions in the numerator, find a common denominator and simplify:
$$f'(x) = \lim_{h\rightarrow 0} \frac{(x+h)(x+1)-(x)(x+h+1)}{h(x+h+1)(x+1)}$$
04
Expand and simplify the numerator
Expand and simplify the expression in the numerator:
$$f'(x) = \lim_{h\rightarrow 0} \frac{x^2-xh+x+2h-x^2-hx-xh-h^2}{h(x+h+1)(x+1)}$$
$$ f'(x) = \lim_{h\rightarrow 0} \frac{-2xh - h^2 + 2h}{h(x+h+1)(x+1)} $$
05
Factor the numerator and cancel terms
Factor out \(h\) from the numerator and simplify:
$$ f'(x) = \lim_{h\rightarrow 0} \frac{-h(2x + h - 2)}{h(x+h+1)(x+1)} $$
Now cancel out the \(h\):
$$ f'(x) = \lim_{h\rightarrow 0} \frac{2x + h - 2}{(x+h+1)(x+1)} $$
06
Take the limit as \(h\) approaches \(0\)
Take the limit as \(h\) approaches \(0\):
$$ f'(x) = \frac{2x - 2}{(x+1)(x+1)} $$
$$ f'(x) = \frac{2(x-1)}{(x+1)^2} $$
07
Evaluate the derivative at \(P(-2,2)\)
Now we can find the slope of the tangent line at the point \(P(-2,2)\) by plugging in \(x=-2\):
$$m = f'(-2) = \frac{2(-2-1)}{(-2+1)^2} = \frac{-6}{1} = -6$$
08
Determine the equation of the tangent line
With the slope \(m = -6\) and the point \(P(-2,2)\), we can use point-slope form to find the equation of the tangent line:
$$ y - 2 = -6(x+2) $$
Now, simplify the equation:
$$ y = -6x -12 + 2$$
$$ y = -6x -10 $$
So, the equation of the tangent line is: $$ y = -6x - 10 $$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
In mathematics, the derivative of a function at a given point is a fundamental concept that describes how the function is changing at that particular point. The derivative measures the rate at which a function's value is changing. In more intuitive terms, it can be thought of as finding the slope of the tangent line to the function at a specific point.
The derivative is symbolically represented as \( f'(x) \), and it's computed using the limit definition:
Understanding the derivative is essential for solving and interpreting various mathematical problems, such as computing tangent lines, as it enables a precise measurement of instantaneous rates of change at any point of the function.
The derivative is symbolically represented as \( f'(x) \), and it's computed using the limit definition:
- \( f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} \)
Understanding the derivative is essential for solving and interpreting various mathematical problems, such as computing tangent lines, as it enables a precise measurement of instantaneous rates of change at any point of the function.
Limit of a Function
Limits are a key concept when it comes to understanding derivatives. They help us to deal with quantities that approach a certain value, even if they don't actually reach that value. In the context of derivatives, a limit is used to define the slope of a function as the values become infinitesimally close.
To compute the derivative, we consider the limit of the difference quotient as \( h \to 0 \). This step involves:
To compute the derivative, we consider the limit of the difference quotient as \( h \to 0 \). This step involves:
- Taking the function's value at \( x+h \) and \( x \), then finding their difference.
- Dividing this difference by \( h \), the change in \( x \).
- Finding the limit of this expression as \( h \to 0 \) to refine the slope measurement.
Point-Slope Form
The point-slope form is a powerful tool in geometry and algebra for writing an equation of a line when given a point and the slope of that line. The form is expressed as:
Once the slope of the tangent line is known, and with a point \( P \) through which the line passes, a linear equation can easily be derived using the point-slope form. This equation describes the tangent line to a curve at that point and offers an exact linear approximation to the curve in the vicinity of that point. For instance, in the solved exercise, given slope \( m = -6 \) and point \( (-2, 2) \), we derived:
- \( y - y_1 = m(x - x_1) \)
Once the slope of the tangent line is known, and with a point \( P \) through which the line passes, a linear equation can easily be derived using the point-slope form. This equation describes the tangent line to a curve at that point and offers an exact linear approximation to the curve in the vicinity of that point. For instance, in the solved exercise, given slope \( m = -6 \) and point \( (-2, 2) \), we derived:
- \( y - 2 = -6(x + 2) \)
- \( y = -6x - 10 \)