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Chapter 4: Problem 14

Locate the critical points and identify which critical points are stationary points. $$ f(x)=\sin |x| $$

Short Answer

Expert verified
Critical points: \( x = 0, \\ (2k+1) \\ \frac{\pi}{2} \). Stationary points: \( (2k+1) \\ \frac{\pi}{2} \).

Step by step solution

01

Understanding Critical Points

Critical points occur where the derivative of the function is zero or undefined. To find the critical points of the function \( f(x) = \ \sin |x| \), we first need to determine its derivative.
02

Expressing the Function Piecewise

The function \( f(x) = \ \sin |x| \) can be expressed as \( \sin(x) \) for \( x \geq 0 \) and \( \sin(-x) \) for \( x < 0 \) because the absolute value operation makes negative values positive. Thus, the function is piecewise defined.
03

Finding the Derivative of the Function

For \( x \geq 0 \), the derivative is \( f'(x) = \ \cos(x) \). For \( x < 0 \), using the chain rule, \( f'(x) = \ \cos(-x) (-1) = -\ \cos(x) \). Now, we check these derivatives at different points to find where they are zero or undefined.
04

Evaluating Zero Derivatives

Set \( \cos(x) = 0 \) for \( x \geq 0 \), which gives solutions at \( x = \ \frac{\pi}{2}, \ \frac{3\pi}{2}, \ \frac{5\pi}{2}, \ldots \). Similarly, for \( x < 0 \), solve \( -\ \cos(x) = 0 \), which also yields solutions at \( x = -\ \frac{\pi}{2}, -\ \frac{3\pi}{2}, -\ \frac{5\pi}{2}, \ldots \).
05

Checking Undefined Derivatives

The derivative of \( f(x) \) at \( x = 0 \) is undefined because the absolute value introduces a discontinuity in the derivative across \( x = 0 \). Therefore, \( x = 0 \) is a critical point where the derivative is undefined.
06

Identify Stationary Points

Stationary points occur where the derivative is zero. Evaluating from our results, the stationary points are \( x = \ \frac{\pi}{2}, -\ \frac{\pi}{2} \), and any other \( x = \ (2k+1) \ \frac{\pi}{2} \) where \( k \) is an integer.

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Key Concepts

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

Derivative
Understanding the concept of a derivative is essential for identifying critical points in a function. A derivative, simply put, measures how a function changes as its input changes. For example, it can tell us how quickly a function is rising or falling at any given point. This rate of change is pivotal in calculus because it helps us analyze the behavior of graphs and functions.

When calculating the derivative of a function, we're looking for the slope of the tangent line at any point on the graph. If the slope is zero, it indicates a flat area on the graph, which is a potential location for a critical point.

In the given exercise, to find the critical points of the function, we differentiate the piecewise defined function, taking the derivative on each piece separately. This step often reveals where potential turning points occur.
Stationary Points
Stationary points are specific types of critical points where the derivative of a function equals zero. These points on the graph indicate where the function has a horizontal tangent line. Finding these points helps in understanding where the function might have maximum, minimum, or inflection points.
Remember that finding stationary points involves setting the derivative to zero and solving for the variable. In our scenario, the solution of the derivative equation at these stationary points reveals flatness of the graph, meaning no increase or decrease at that particular point. Therefore, identifying points like \( x = \frac{\pi}{2} \) or \( x = -\frac{\pi}{2} \) is crucial because they demonstrate where the function changes behavior.

Overall, while stationary points indicate where the derivative is zero, they might not always be the point of highest or lowest values, they just provide a snapshot of potential interest regarding the function's behavior.
Piecewise Function
A piecewise function is a type of function defined by different expressions depending on the value of the input. These functions split their definition over different intervals of the domain, specifying separate expressions for each.

In the case of \( f(x) = \sin |x| \), this function was expressed piecewise because the absolute value changes the expression based on whether \( x \) is positive or negative. So, \( \sin(x) \) is used when \( x \geq 0 \), and \( \sin(-x) \) is used when \( x < 0 \).

This piecewise representation is necessary because the absolute function alters the input before the trigonometric function is applied. By splitting up the expressions, we ensure the derivative correctly addresses the behavior of the function at every part of the domain. Understanding this allows us to systematically address changes across a function's domain, enhancing comprehension of its graph's overall behavior.

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Most popular questions from this chapter

The function \(s(t)\) describes the position of a particle moving along a coordinate line, where \(s\) is in feet and \(t\) is in seconds. (a) Find the velocity and acceleration functions. (b) Find the position, velocity, speced, and acceleration at time \(t=1 .\) (c) At what times is the particle stopped? (d) When is the particle speeding up? Slowing down? (e) Find the total distance traveled by the particle from time \(t=0\) to time \(t=5\). $$ s(t)=\frac{1}{4} t^{2}-\ln (t+1), \quad t \geq 0 $$

An annuity is a sequence of equal payments that are paid or received at regular time intervals. For example, you may want to deposit equal amounts at the end of each year into an interest-bearing account for the purpose of accumulating a lump sum at some future time. If, at the end of each year, interest of \(i \times 100 \%\) on the account balance for that year is added to the account, then the account is said to pay \(i \times 100 \%\) interest, compounded annually. It can be shown that if payments of \(Q\) dollars are deposited at the end ofeach year into an account that pays \(i \times 100 \%\) compounded annually, then at the time when the \(n\) th payment and the accrued interest for the past year are deposited, the amount \(S(n)\) in the account is given by the formula $$S(n)=\frac{Q}{i}\left[(1+i)^{n}-1\right]$$ Suppose that you can invest \(\$ 5000\) in an interest-bearing account at the end of each year, and your objective is to have \(\$ 250,000\) on the 25th payment. Approximately what annual compound interest rate must the account pay for you to achieve your goal? [Hint: Show that the interest rate \(i\) satisfies the equation \(50 i=(1+i)^{25}-1,\) and solve it using Newton's Method.]

Let \(g(x)=x e^{x}-e^{x} .\) Find \(f(x)\) so that \(f^{\prime}(x)=g^{\prime}(x)\) and \(f(1)=2\)

Verify that the hypotheses of the Mean-Value Theorem are satisfied on the given interval, and find all values of \(c\) in that interval that satisfy the conclusion of the theorem. $$ f(x)=x^{3}+x-4 ;[-1,2] $$

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