/*! 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 31 Derivatives and tangent lines ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 31

Derivatives and tangent lines a. For the following functions and values of \(a,\) find \(f^{\prime}(a)\) b. Determine an equation of the line tangent to the graph of \(f\) at the point \((a, f(a))\) for the given value of \(a\) $$f(x)=\frac{1}{\sqrt{x}} ; a=\frac{1}{4}$$

Short Answer

Expert verified
Answer: The equation of the tangent line to the graph of the function at the given point is $$y = -2x + \frac{5}{2}$$.

Step by step solution

01

Find the derivative of the function

To find the derivative of $$f(x)=\frac{1}{\sqrt{x}}$$, we can rewrite the function as $$f(x)=x^{-\frac{1}{2}}$$. Now, using the power rule, the derivative is given by: $$f^{\prime}(x)=-\frac{1}{2}x^{-\frac{3}{2}}$$
02

Compute the derivative at the given point

We are given the value of $$a=\frac{1}{4}$$ and asked to find $$f^{\prime}(a)$$. Using the derivative found in step 1, we have: $$f^{\prime}\left(\frac{1}{4}\right)=-\frac{1}{2}\left(\frac{1}{4}\right)^{-\frac{3}{2}}$$ Calculating the derivative at the given point, we get: $$f^{\prime}\left(\frac{1}{4}\right)=-2$$
03

Find the coordinates of the given point

We need to find the point $$(a, f(a))$$ on the graph of the function. Since we are given the value of $$a=\frac{1}{4}$$, we can plug it in the original function to get: $$f\left(\frac{1}{4}\right)=\frac{1}{\sqrt{\frac{1}{4}}}$$ Calculating, we get: $$f\left(\frac{1}{4}\right)=2$$ So, the point is $$\left(\frac{1}{4}, 2\right)$$.
04

Determine the equation of the tangent line

Using the point-slope form of a linear equation, and the slope of the tangent line being $$m=-2$$ as found in step 2, we can find the equation of the tangent line: $$y - f(a) = f^{\prime}(a) (x - a)$$ Plugging in the values we found earlier, we get: $$y - 2 = -2\left(x - \frac{1}{4}\right)$$ Now, simplify and rewrite it in the slope-intercept form: $$y = -2x + 2\left(\frac{1}{4}\right) + 2$$ $$y = -2x + \frac{5}{2}$$ So, the equation of the tangent line to the graph of the function at the given point is $$y = -2x + \frac{5}{2}$$.

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!

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

Assuming the first and second derivatives of \(f\) and \(g\) exist at \(x\), find a formula for \(\frac{d^{2}}{d x^{2}}(f(x) g(x))\)

Two boats leave a port at the same time, one traveling west at \(20 \mathrm{mi} / \mathrm{hr}\) and the other traveling southwest at \(15 \mathrm{mi} / \mathrm{hr} .\) At what rate is the distance between them changing 30 min after they leave the port?

Let \(f\) and \(g\) be differentiable functions with \(h(x)=f(g(x)) .\) For a given constant \(a,\) let \(u=g(a)\) and \(v=g(x),\) and define $$H(v)=\left\\{\begin{array}{ll} \frac{f(v)-f(u)}{v-u}-f^{\prime}(u) & \text { if } v \neq u \\ 0 & \text { if } v=u. \end{array}\right.$$ a. Show that \(\lim _{x \rightarrow u} H(v)=0\) b. For any value of \(u\) show that $$f(v)-f(u)=\left(H(v)+f^{\prime}(u)\right)(v-u).$$ c. Show that. $$h^{\prime}(a)=\lim _{x \rightarrow a}\left(\left(H(g(x))+f^{\prime}(g(a))\right) \cdot \frac{g(x)-g(a)}{x-a}\right).$$ d. Show that \(h^{\prime}(a)=f^{\prime}(g(a)) g^{\prime}(a)\).

Derivative of \(u(x)^{v(x)}\) Use logarithmic differentiation to prove that $$\frac{d}{d x}\left(u(x)^{v(x)}\right)=u(x)^{v(x)}\left(\frac{d v}{d x} \ln u(x)+\frac{v(x)}{u(x)} \frac{d u}{d x}\right)$$.

A spring hangs from the ceiling at equilibrium with a mass attached to its end. Suppose you pull downward on the mass and release it 10 inches below its equilibrium position with an upward push. The distance \(x\) (in inches) of the mass from its equilibrium position after \(t\) seconds is given by the function \(x(t)=10 \sin t-10 \cos t,\) where \(x\) is positive when the mass is above the equilibrium position. a. Graph and interpret this function. b. Find \(\frac{d x}{d t}\) and interpret the meaning of this derivative. c. At what times is the velocity of the mass zero? d. The function given here is a model for the motion of an object on a spring. In what ways is this model unrealistic?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

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.