/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Given that \(f(3)=-1\) and \(f^{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 7

Given that \(f(3)=-1\) and \(f^{\prime}(3)=5\), find an equation for the tangent line to the graph of \(y=f(x)\) at \(x=3\).

Short Answer

Expert verified
The equation of the tangent line is \(y = 5x - 16\).

Step by step solution

01

Understand the Problem

To find the equation of the tangent line to the curve \(y=f(x)\) at a specific point \(x=3\), we need a point on the line and the slope of the line at that point.
02

Identify Known Values

From the problem, we know two critical values: \(f(3) = -1\), meaning the point where the tangent line touches the curve is \((3, -1)\), and \(f'(3) = 5\), which is the slope of the tangent line at \(x=3\).
03

Use Point-Slope Formula

The point-slope form of a line is given by the formula \(y - y_1 = m(x - x_1)\), where \((x_1,y_1)\) is a point on the line, and \(m\) is the slope. Here, \(x_1 = 3\), \(y_1 = -1\), and \(m = 5\).
04

Substitute Values into the Equation

Substitute the known values into the point-slope equation: \(y - (-1) = 5(x - 3)\). Simplify it to get the equation: \(y + 1 = 5(x - 3)\).
05

Simplify the Equation

Reorganize the equation into slope-intercept form (\(y = mx + b\)) or another preferred form: Solve \(y + 1 = 5(x - 3)\) to get \(y = 5x - 15 - 1\), which simplifies to \(y = 5x - 16\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Point-Slope Form
The point-slope form is a very handy way to write the equation of a line. It's particularly useful when you know a single point on the line and the slope of the line. This form is expressed as:
  • \[ y - y_1 = m(x - x_1) \]
  • In this formula, \((x_1, y_1)\) is the point with coordinates on the line.
  • \(m\) represents the slope, which shows how steep the line is.
To apply this form, plug in the point you know and the slope. It's very direct and saves you from needing to find extra points.
Using this form is like building the bridge between a point and the slope, which gives you a clear path to finding the equation of the line. This approach is straightforward for writing down tangent lines, as it directly relates to the change in your function at a specific point.
Derivative
Derivatives are a core concept in calculus and serve as a measure of how a function is changing at any given point. Essentially, the derivative provides us with the function's slope at any location.
When finding a tangent line to a curve, the derivative is crucial. This is because the derivative at a specific point gives the exact slope of the tangent line there. In our context, with \(f'(3) = 5\):
  • The value \(5\) is the slope of the tangent line at \(x=3\).
  • This reflects how the function \(f(x)\) behaves at that precise location.
Knowing this slope is like having the angle at which you tilt your line to just touch the curve. It's precise, and that's what makes derivatives so powerful. They provide the rate of change, which is essential for understanding the behavior of functions.
Slope-Intercept Form
The slope-intercept form is another popular way to express the equation of a line. It is expressed as:
  • \[ y = mx + b \]
  • Here, \(m\) stands for the slope, giving the steepness of the line.
  • \(b\) is the y-intercept, which is the point where the line crosses the y-axis.
When you transform a line's equation into slope-intercept form, it becomes very easy to see both how steep the line is and where it starts on the y-axis.
In our case, after simplifying from the point-slope form, we found:
  • \[ y = 5x - 16 \]
  • Here, the slope \(m\) is \(5\), indicating the line is going upwards fairly quickly.
  • The y-intercept \(b\) is \(-16\), showing where the line crosses the y-axis.
This form is concise and visually intuitive, making it easier to graph the line or integrate it into larger problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that if \(x \neq 0\), then \(y=1 / x\) satisfies the equation \(x^{3} y^{\prime \prime}+x^{2} y^{\prime}-x y=0\)

Find \(d y / d x\) \(y=\cos ^{3}(\sin 2 x)\)

Find a function \(y=a x^{2}+b x+c\) whose graph has an \(x\) -intercept of 1 , a \(y\) -intercept of \(-2\), and a tangent line with a slope of \(-1\) at the \(y\) -intercept.

Find the \(x\) -coordinate of the point on the graph of \(y=x^{2}\) where the tangent line is parallel to the secant line that cuts the curve at \(x=-1\) and \(x=2\).

Determine whether the statement is true or false. Explain your answer. $$ \begin{aligned} &\text { If } f^{\prime}(2)=5 \text { , then } \\ &\left.\frac{d}{d x}\left[4 f(x)+x^{3}\right]\right|_{x=2}=\left.\frac{d}{d x}[4 f(x)+8]\right|_{x=2}=4 f^{\prime}(2)=20 \end{aligned} $$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.