/*! 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 6 Find the slope of the tangent li... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the slope of the tangent line to the graph of the function at the given point. \(g(x)=\frac{3}{2} x+1, \quad(-2,-2)\)

Short Answer

Expert verified
The slope of the tangent line to the function \(g(x)=\frac{3}{2} x+1\) at the point (-2,-2) is \(\frac{3}{2}\).

Step by step solution

01

Identify the slope of the function

For the function \(g(x)=\frac{3}{2} x+1\), the slope (which is the physical interpretation of the derivative for a linear function) is \(\frac{3}{2}\) which is the coefficient of \(x\). This is the same regardless of the point on the line, as the function is linear.
02

Confirm the point lies on the function

Before claiming the slope of the tangent line at the point (-2,-2), check to ensure this point actually lies on the function. Substitute the coordinates of the point into the original function. With \(x=-2\), the function's value becomes \(\frac{3}{2}*(-2)+1=-3+1=-2\). Since the result is -2 which equals to the y-coordinate of the given point, the point (-2,-2) does indeed lie on the function.
03

State the slope of the tangent line at the given point

The slope of the tangent line at any point on the function \(g(x)\) equals the derivative which is the coefficient of \(x\), and that derivative or slope is \(\frac{3}{2}\)

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 91Ó°ÊÓ!

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

Linear and Quadratic Approximations The linear and quadratic approximations of a function \(f\) at \(x=a\) are \(P_{1}(x)=f^{\prime}(a)(x-a)+f(a)\) and \(P_{2}(x)=\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}+f^{\prime}(a)(x-a)+f(a)\) \(\begin{array}{llll}\text { In Exercises } & 133-136, & \text { (a) find the specified linear and }\end{array}\) quadratic approximations of \(f,\) (b) use a graphing utility to graph \(f\) and the approximations, (c) determine whether \(P_{1}\) or \(P_{2}\) is the better approximation, and (d) state how the accuracy changes as you move farther from \(x=a\). $$ \begin{array}{l} f(x)=e^{-x^{2} / 2} \\ a=0 \end{array} $$

In Exercises \(81-88\), (a) find an equation of the tangent line to the graph of \(f\) at the indicated point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. \(\frac{\text { Function }}{y=2 e^{1-x^{2}}} \quad \frac{\text { Point }}{\left(1,2\right)}\)

Find equations of both tangent lines to the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\) that passes through the point (4,0).

A television camera at ground level is filming the lift-off of a space shuttle at a point 750 meters from the launch pad. Let \(\theta\) be the angle of elevation of the shuttle and let \(s\) be the distance between the camera and the shuttle (as shown in the figure). Write \(\theta\) as a function of \(s\) for the period of time when the shuttle is moving vertically. Differentiate the result to find \(d \theta / d t\) in terms of \(s\) and \(d s / d t\).

Let \(L\) be any tangent line to the curve \(\sqrt{x}+\sqrt{y}=\sqrt{c}\). Show that the sum of the \(x\) - and \(y\) -intercepts of \(L\) is \(c\).

See all solutions

Recommended explanations on Math Textbooks

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.