/*! 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 46 Find an equation of the line \(/... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 46

Find an equation of the line \(/\) tangent to the graph of \(f\) at the given point. $$ f(x)=\sqrt{x} ;(16,4) $$

Short Answer

Expert verified
The tangent line equation is \(y = \frac{1}{8}x + 2\).

Step by step solution

01

Differentiate the Function

To find the tangent line, we need the derivative of the function because the derivative at a point gives the slope of the tangent. Given the function \(f(x) = \sqrt{x} \), rewrite it as \(f(x) = x^{1/2}\). Differentiate \(f(x)\) to find \(f'(x)\):\[ f'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \]
02

Evaluate the Derivative at Given Point

Substitute \(x = 16\) into the derivative to find the slope of the tangent line at \((16, 4)\):\[ f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{2 \times 4} = \frac{1}{8} \]
03

Use Point-Slope Form of a Line

With the slope \(m = \frac{1}{8}\) and the point \((16, 4)\), we use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), to find the equation of the tangent line:\[ y - 4 = \frac{1}{8}(x - 16) \]
04

Simplify the Equation of the Tangent Line

Simplify the equation to put it in slope-intercept form \(y = mx + b\):\[ y - 4 = \frac{1}{8}x - 2 \ y = \frac{1}{8}x + 2 \]
05

Conclusion: Equation of the Tangent Line

The equation of the line tangent to \(f(x) = \sqrt{x}\) at the point \((16, 4)\) is \(y = \frac{1}{8}x + 2\).

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.

Tangent Line
A tangent line is a straight line that just "touches" a curve at a particular point, without cutting across the curve. Think of it as the line that best approximates the curve at that point. It provides a snapshot of the curve's direction. For a given function, such as \( f(x) = \sqrt{x} \), the tangent line at a specific point provides the best linear approximation of the function near that point.
  • In our case, the point is \((16, 4)\).
  • The tangent line visually represents how the function behaves precisely at the point \((16, 4)\).
This tangent line helps us understand the instantaneous rate of change or the precise slope at the exact point of tangency.
Derivative
The derivative of a function measures how the function's output (or \( y \)-value) changes as its input (or \( x \)-value) changes. It is essentially the function's rate of change or slope at each point.
  • The notation used for the derivative is \( f'(x) \).
  • For \( f(x) = \sqrt{x} \), the derivative \( f'(x) \) is \( \frac{1}{2\sqrt{x}} \).
To find the equation of the tangent line, we first find the derivative to understand how steep the curve is at a specific point. This step is crucial in the process because it gives us the slope of the tangent line at the desired point.
Slope
The slope of a line is a measure of its steepness. It is calculated as the "rise over run," that is, the change in \( y \) over the change in \( x \). For tangent lines, the slope at a specific point on a curve is found by evaluating the derivative at that point.
  • The slope of the tangent line for our function at \((16, 4)\) is \( \frac{1}{8} \).
By substituting \( x = 16 \) into the derivative \( f'(x) = \frac{1}{2\sqrt{x}} \), we found that the slope is \( \frac{1}{8} \). This slope tells us exactly how quickly \( f(x) \) is changing at \( x = 16 \).
Point-Slope Form
The point-slope form is a linear equation format used to describe the equation of a line when a point on the line and the slope are known. The formula is:\[ y - y_1 = m(x - x_1) \]where \((x_1, y_1)\) is a point on the line, and \( m \) is the slope of the line.
  • For the point \((16, 4)\) with a slope of \( \frac{1}{8} \), the equation becomes \( y - 4 = \frac{1}{8}(x - 16) \).
Using the point-slope form is an effective way to rapidly find the equation of the tangent line once the slope and a single point are known. By rearranging the equation into a more familiar slope-intercept form \( y = mx + b \), we get \( y = \frac{1}{8}x + 2 \), making it easy to graph or use in further calculations.

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

$$ \lim _{x \rightarrow 0^{+}} \frac{1}{1-e^{\sqrt{x}}} $$

Decide which of the given one-sided or twosided limits exist as numbers, which as \(\infty\), which as \(-\infty\), and which do not exist. Where the limit is a number, evaluate it. $$ \lim _{x \rightarrow 0} \sqrt{\frac{1}{x^{2}}} $$

Decide which of the given one-sided or two-sided limits exist as numbers, which as \(\infty\), which as \(-\infty\), and which do not exist. Where the limit is a number, evaluate it. $$ \lim _{x \rightarrow 0} \frac{x}{|x|} $$

Solve the inequality for \(x\). $$ (x+1)(x-2)\left(2 x-\frac{1}{4}\right) \geq 0 $$

Decide which of the given one-sided or twosided limits exist as numbers, which as \(\infty\), which as \(-\infty\), and which do not exist. Where the limit is a number, evaluate it. $$ \lim _{x \rightarrow 1^{+}}\left(\sqrt{x^{2}-x}+x\right) $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

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.