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Chapter 3: Problem 15

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

Short Answer

Expert verified
Answer: The equation of the tangent line is y = x.

Step by step solution

01

Find the derivative of the function f(x)

First, we need to find the derivative of the function \(f(x)\) using the definition of a derivative: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ Here, the function is \(f(x) = \frac{1}{x}\), so we need to find \(f(x + h)\) first: $$f(x + h) = \frac{1}{x + h}$$ Now, let's plug in \(f(x)\) and \(f(x + h)\) into the derivative formula and simplify: $$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h}$$ $$f'(x) = \lim_{h \to 0} \frac{x - (x+h)}{x(x+h)h}$$ $$f'(x) = \lim_{h \to 0} \frac{-h}{x(x+h)h}$$ $$f'(x) = \lim_{h \to 0} \frac{-1}{x(x+h)}$$
02

Evaluate the derivative at P

Now, let's find the slope, \(m\), of the tangent line to the function \(f\) at point \(P\) with coordinates \((-1, -1)\) by evaluating the derivative at \(x = -1\): $$m = f'(-1) = \frac{-1}{-1(-1+0)} = \frac{-1}{-1} = 1$$ So, the slope of the tangent line is \(1\). #b. Finding the equation of the tangent line#
03

Write the equation of the tangent line using the point-slope form

Now that we have the slope of the tangent line and the coordinates of the point \(P\), we can write the equation of the tangent line using the point-slope form of a linear equation, which is \(y - y_1 = m(x - x_1)\): Let \((x_1, y_1) = (-1, -1)\) and \(m = 1\): $$y - (-1) = 1(x - (-1))$$ $$y + 1 = x + 1$$ $$y = x$$ So, the equation of the tangent line at point \(P\) is \(y = x\). #c. Plotting the graph of the function and tangent line#
04

Sketch the graph of f(x) and the tangent line at P

To sketch the graph of the function \(f(x) = \frac{1}{x}\) and the tangent line \(y = x\) at point \(P\), follow these steps: 1. Draw the coordinate axes and label them. 2. Plot the point \(P(-1, -1)\) and label it. 3. Sketch the graph of \(f(x) = \frac{1}{x}\), which is a hyperbola. 4. Sketch the tangent line with the equation \(y = x\). Now, you should have a graph that shows the function \(f(x) = \frac{1}{x}\) with its tangent line at the point \(P(-1, -1)\), which has the equation \(y = x\).

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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 is changing at any given point. It's a fundamental concept in calculus and provides us the slope of the tangent line to the curve at a particular point.
The derivative is often denoted by \(f'(x)\) or \(\frac{dy}{dx}\), and it's a powerful tool because it can tell us not just how fast a function changes, but also the behavior of that function.
For instance, if the derivative is positive at a point, then the function is increasing at that point; if it's negative, the function is decreasing. This makes derivatives very useful in various real-world contexts, like physics and economics, where understanding change is crucial.
Limit Definition
The limit definition of a derivative is one of the foundational ideas in calculus. It's expressed as: \[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\] This equation describes the derivative as a limit, measuring how \(f(x)\) changes as we make a small increment \(h\) approaching zero.
By using this approach, we calculate the instantaneous rate of change of the function, which is the slope of the tangent line at a particular point.
This method can seem complex at first, but it essentially boils down to finding how the function behaves as it approaches a very small interval of change. This is how we formally define and calculate derivatives when given any function.
Slope
Slope is a measure of how steep a line is. For a straight line, slope is calculated as the rise over run, or \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
In the context of a curve, the slope of the tangent line at a precise point gives us valuable information about how the function behaves at that spot.
When we find the derivative of a function and evaluate it at a certain point, what we are actually doing is determining the slope of the tangent line to the curve at that point.
  • If the slope is positive, the function is increasing.
  • If the slope is negative, it is decreasing.
  • If the slope is zero, the tangent line is flat, and it might suggest a local maximum or minimum.
Point-Slope Form
The point-slope form is a technique used to find the equation of a line, given its slope and a point on the line. The formula is: \(y - y_1 = m(x - x_1)\) Here, \((x_1, y_1)\) is a point on the line, and \(m\) is the slope.
This form is particularly useful when dealing with tangent lines in calculus because it allows us to easily write an equation for a tangent line as soon as we know a point and a slope.
For example, having determined that at point \((-1, -1)\) the slope of the tangent line is 1, the equation of the tangent line is found using:
  • Replace \(x_1\) and \(y_1\) with -1.
  • Replace \(m\) with 1.
Thus, the equation becomes \(y = 1(x + 1) - 1\), simplifying to \(y = x\).

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Most popular questions from this chapter

Electrostatic force The magnitude of the electrostatic force between two point charges \(Q\) and \(q\) of the same sign is given by \(F(x)=\frac{k Q q}{x^{2}},\) where \(x\) is the distance (measured in meters) between the charges and \(k=9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\) is a physical constant (C stands for coulomb, the unit of charge; N stands for newton, the unit of force). a. Find the instantaneous rate of change of the force with respect to the distance between the charges. b. For two identical charges with \(Q=q=1 \mathrm{C},\) what is the instantaneous rate of change of the force at a separation of \(x=0.001 \mathrm{m} ?\) c. Does the magnitude of the instantaneous rate of change of the force increase or decrease with the separation? Explain.

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals. . A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Use implicit differentiation if needed to find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(x y=a ; x^{2}-y^{2}=b,\) where \(a\) and \(b\) are constants

A rope passing through a capstan on a dock is attached to a boat offshore. The rope is pulled in at a constant rate of \(3 \mathrm{ft} / \mathrm{s}\) and the capstan is \(5 \mathrm{ft}\) vertically above the water. How fast is the boat traveling when it is \(10 \mathrm{ft}\) from the dock?

Graphing \(f\) and \(f^{\prime}\) a. Graph \(f\) with a graphing utility. b. Compute and graph \(f^{\prime}\) c. Verify that the zeros of \(f^{\prime}\) correspond to points at which \(f\) has \(a\) horizontal tangent line. $$f(x)=(x-1) \sin ^{-1} x \text { on }[-1,1]$$

Vertical tangent lines a. Determine the points where the curve \(x+y^{2}-y=1\) has a vertical tangent line (see Exercise 53 ). b. Does the curve have any horizontal tangent lines? Explain.
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