Question

Suppose that the function \(\mathrm{f}\) is defined for all \(\mathrm{x}\) such that \(|\mathrm{x}|>1\) and has the property that \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x} \sqrt{\mathrm{x}^{2}-1}}\) for all such \(\mathrm{x}\). (a) Explain why there exists two constants \(\mathrm{A}\) and \(\mathrm{B}\) such that \(f(x)=\sec ^{-1} x+A\) if \(x>1\) \(f(x)=-\sec ^{-1} x+B\) if \(x<-1\) (b) Determine the values of \(\mathrm{A}\) and \(\mathrm{B}\) so that \(f(2)=1=\mathrm{f}(-2)\). Then sketch the graph of \(y=f(x)\)

Short answer

#tag_title#Short Answer#tag_content# There exist constants A and B such that the given function descriptions are true since the derivatives of both sec⁻¹x and -sec⁻¹x match the given derivative f'(x). We find that A = 1 - π/3 and B = 1 + 2π/3. The graph of y=f(x) consists of two cases: (1) for x > 1, it follows y = sec⁻¹x + 1 - π/3 with a vertical asymptote at x = 1, and (2) for x < -1, it follows y=-sec⁻¹x + 1 + 2π/3 with a vertical asymptote at x = -1.

Step-by-step solution

Verify the derivatives

Let's first check the derivative of \(\sec^{-1}x\) and \(-\sec^{-1}x\) for \(x>1\) and \(x<-1\) respectively. Using chain rule, we have: \(\frac{d(\sec^{-1}x)}{dx}=\frac{1}{\sqrt{1-x^{-2}}}\cdot(-\frac{1}{x^2})=\frac{1}{x\sqrt{x^2-1}}\) \(\frac{d(-\sec^{-1}x)}{dx}=-\frac{1}{\sqrt{1-x^{-2}}}\cdot(-\frac{1}{x^2})=\frac{1}{x\sqrt{x^2-1}}\) Notice that the derivatives of both \(\sec^{-1}x\) and \(-\sec^{-1}x\) match the given derivative \(f'(x)\). Adding a constant \(A\) or \(B\) to these functions will not change the derivative, as the derivative of a constant is zero. Now, we can conclude that there exist constants \(A\) and \(B\) such that \(f(x)=\sec^{-1}x+A\) when \(x>1\) and \(f(x)=-\sec^{-1}x+B\) when \(x<-1\).

Determine the values of A and B

Given that \(f(2)=1\) and \(f(-2)=1\), we can use these conditions to solve for the constants \(A\) and \(B\). For \(f(2)=1\), we know that \(x>1\), so we will use the expression: \(f(x)=\sec^{-1}x+A\): \(1=\sec^{-1}(2)+A\) To find \(\sec^{-1}(2)\), we evaluate \(\arccos(\frac{1}{2})=\frac{\pi}{3}\). So, \(1=\frac{\pi}{3}+A\). Thus, \(A=1-\frac{\pi}{3}\). For \(f(-2)=1\), we know that \(x<-1\), so we will use the expression: \(f(x)=-\sec^{-1}x+B\): \(1=-\sec^{-1}(-2)+B\) Now, we have \(\sec^{-1}(-2) = \arccos(-\frac{1}{2})=\frac{2\pi}{3}\). So, \(1=-\frac{2\pi}{3}+B\). Thus, \(B=1+\frac{2\pi}{3}\).

Sketch the graph of y=f(x)

To sketch the graph of \(y=f(x)\), we need to consider the two cases: when \(x>1\) and when \(x<-1\). When \(x>1\), the graph follows \(y=\sec^{-1}x+1-\frac{\pi}{3}\), which is the graph of the inverse secant function that has been shifted up by \((1-\frac{\pi}{3})\) units. At \(x=1\), there is a vertical asymptote. Similarly, when \(x<-1\), the graph follows \(y=-\sec^{-1}x+1+\frac{2\pi}{3}\), which is the graph of the inverse secant function reflected across the x-axis and shifted up by \((1+\frac{2\pi}{3})\) units. At \(x=-1\), there is also a vertical asymptote. With these transformations, you will be able to sketch the graph of \(y=f(x)\).

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of the trigonometric functions and are essential in calculus. They provide the angles associated with a given trigonometric value. One common inverse trigonometric function is the arcsecant, often written as \(\sec^{-1}x\) or \(\arccos(\frac{1}{x})\). Like their trigonometric counterparts, inverse trigonometric functions have specific domains and ranges that allow for unique values.

For instance, \(\sec^{-1}x\) is defined for \(x > 1\) or \(x < -1\), which corresponds to the original function \(\sec(\theta)\) being defined only for certain values of \(\theta\) to ensure that each \(x\) value has only one \(\theta\) (angle) associated with it. Understanding the behavior of these functions, such as their asymptotes and reflective properties, is key in solving many calculus problems, such as integrals involving trigonometric expressions.

Derivatives

Derivatives are a cornerstone of calculus, representing the rate of change of a function with respect to a variable. The notation \(f'(x)\) or \(\frac{df}{dx}\) denotes the derivative of function \(f\) with respect to \(x\).

When we take the derivative of an inverse trigonometric function like \(\sec^{-1}x\), we use the chain rule to find the relationship between the rates of change in \(x\) and the function itself. The derivative of \(\sec^{-1}x\) is particularly interesting because it is defined in terms of the original variable \(x\) and the square root of an expression involving \(x\), as seen in the step-by-step solution. Knowing the derivatives of inverse trigonometric functions is crucial for solving integration problems where these functions are involved. The ability to correctly compute derivatives underpins a student's success in integral calculus and other advanced mathematical fields.

Graph Sketching

Graph sketching is an analytical skill in mathematics that involves understanding the effects of transformations on function graphs. The process usually begins with the basic graph of a known function and then applies shifts, reflections, and scaling to create the graph of a more complex function.

In calculus, sketching the graph of a function helps to visualize the behavior of the function across its domain. For inverse trigonometric functions such as \(\sec^{-1}x\), recognizing vertical asymptotes, which occur at the points where the original trigonometric function – in this case, \(\sec(x)\) – is undefined, is essential for accurate sketching. In the step-by-step solution provided, we see how the constants \(A\) and \(B\) shift the graph vertically and how the negative sign in front of \(\sec^{-1}x\) for \(x < -1\) leads to a reflection across the x-axis. Such knowledge facilitates not only the visualization of functions but also better comprehension of calculus concepts like limits and integrals.