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Chapter 3: Problem 33

Find \(a\) in each of the following cases. a. \(f(x)=-2 x^{2} ; f^{\prime}(a)=12\) b. \(f(x)=3 x+x^{2} ; f^{\prime}(a)=13\) c. \(f(x)=1 / x ; f^{\prime}(a)=-\frac{1}{9}\) (There are two possible values for \(a\).) d. \(f(x)=\sin x ; f^{\prime}(a)=\sqrt{3} / 2\) (There are infinitely many possible values for \(a\).)

Short Answer

Expert verified
a. \(-3\); b. \(5\); c. \(\pm 3\); d. Infinitely many at intervals of \(2\pi\).

Step by step solution

01

Derive the Function (a)

For the case \(f(x) = -2x^2\), we first need to find its derivative. The power rule tells us that \(f'(x) = -4x\). Set this equal to 12 as given: \(-4a = 12\).
02

Solve for a (a)

To solve \(-4a = 12\), divide both sides by \(-4\) to get \(a = -3\).
03

Derive the Function (b)

For the case \(f(x) = 3x + x^2\), the derivative is found by applying the power rule: \(f'(x) = 3 + 2x\). Set this equal to 13: \(3 + 2a = 13\).
04

Solve for a (b)

To find \(a\), subtract 3 from both sides: \(2a = 10\), and then divide by 2, \(a = 5\).
05

Derive the Function (c)

For the function \(f(x) = 1/x\), use the formula for the derivative of \(x^{-1}\): \(f'(x) = -1/x^2\). Set this equal to \(-\frac{1}{9}\): \(-\frac{1}{a^2} = -\frac{1}{9}\).
06

Solve for a (c)

Solving \(-\frac{1}{a^2} = -\frac{1}{9}\), we equate \(a^2 = 9\). Thus, \(a\) can be \(3\) or \(-3\).
07

Derive the Function (d)

For the function \(f(x) = \sin x\), the derivative is \(f'(x) = \cos x\). Set this equal to \(\sqrt{3}/2\).
08

Solve for a (d)

Values of \(x\) where \(\cos x = \sqrt{3}/2\) are \(x = \frac{\pi}{6} + 2n\pi\) or \(x = \frac{11\pi}{6} + 2n\pi\), where \(n\) is an integer, indicating the infinitude of solutions.

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Key Concepts

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

Derivatives
Derivatives are fundamental in calculus because they tell us about how a function changes. Essentially, the derivative of a function gives us the slope of the tangent to the function at any point. If you think of a graph, the tangent line just "touches" the curve without intersecting it at a single point. Finding this slope is what derivatives do.
  • The Basics: Derivatives measure the rate of change. In simpler terms, it鈥檚 how much one quantity changes when another quantity changes.
  • Why It's Important: Derivatives help us understand how things change, which is crucial in fields like physics, engineering, and economics.
To calculate a derivative, we use various rules and formulas which cater to different types of functions. The Power Rule and the derivatives of trigonometric functions are just a couple of these essential tools.
Critical Points
Critical points are the places on a graph where a function's derivative is zero. These points are where the function could potentially have a maximum, minimum, or an inflection point. At these points, the tangent to the curve is horizontal, meaning the slope is zero.
  • Finding Critical Points: You first find the derivative of a function and then set it equal to zero. Solve the resulting equation to find the critical points.
  • What They Tell Us: Critical points help in understanding the behavior of a graph, such as where it peaks (has a maximum) or dips (has a minimum).
By analyzing critical points, one can understand the dynamic properties of functions and solve practical problems like finding maximum profit or minimum cost.
Trigonometric Functions
Trigonometric functions like sine, cosine, and tangent describe the angles and lengths in right triangles, but their applications go far beyond basic geometry. These functions are periodic, which means they repeat values in regular intervals. In calculus, these functions have unique derivatives that behave differently compared to polynomial functions.
  • Common Derivatives: The derivative of \(\sin x\) is \(\cos x\), and for \(\cos x\), it is \(-\sin x\).
  • Application: They're essential in modeling cyclical phenomena like sound waves and tides, and in computing problems involving circuits in electrical engineering.
Understanding these trigonometric derivatives enhances one's ability to solve physics and engineering problems that involve periodic motion.
Power Rule
The Power Rule is a simple yet powerful tool used for finding derivatives of polynomial functions. It's a shortcut that saves time and effort, especially when dealing with powers of variables.
  • How It Works: For any function of the form \(f(x) = x^n\), the derivative \(f'(x)\) is \(nx^{n-1}\).
  • Example: If \(f(x) = x^3\), applying the power rule gives \(f'(x) = 3x^2\).
The Power Rule is applicable when the function is a polynomial, and it greatly simplifies the problem-solving process by avoiding more complex derivative calculations. It is one of the first and most essential rules you'll learn when tackling derivatives in calculus.

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