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Chapter 5: Problem 72

Find any relative extrema of the function. \(h(x)=\arcsin x-2 \arctan x\)

Short Answer

Expert verified
The function has relative maximum at \(x=-4, 1\), and relative minimum at \(x=-1, 4\)

Step by step solution

01

Finding the Derivative

Find the first derivative of the function \(h(x)=\arcsin x-2 \arctan x\). Use the facts that the derivative of \(\arcsin x\) is \(\frac{1}{\sqrt{1-x^2}}\) and that the derivative of \(\arctan x\) is \(\frac{1}{1+x^2}\). So, the derivative of the function is \(h'(x) = \frac{1}{\sqrt{1-x^2}}-\frac{4}{1+x^2}\).
02

Setting the Derivative to Zero

Set \(h'(x) = 0\) to find critical points. This gives the equation \(0 = \frac{1}{\sqrt{1-x^2}}-\frac{4}{1+x^2}\). Solve this equation for \(x\). Multiplying everything by \(1+x^2\) and \(\sqrt{1-x^2}\) to get rid of the denominators results in \(0 = (1+x^2)\sqrt{1-x^2} - 4\sqrt{1-x^2}\). Grouping like terms, we have \(\sqrt{1-x^2} = 4 - x^2\). Squaring both sides gives \(1 - x^2 = 16 -8x^2 + x^4 \), or \(0 = x^4 - 9x^2 + 16\).
03

Solve the Quadratic Equation

The equation \(0 = x^4 - 9x^2 + 16\) is quadractic in term of \(x^2\). Let \(y = x^2\), we then have \(0 = y^2 - 9y + 16\). Solving for \(y\) gives \(y_1 = 1, y_2 = 16\). Therefore, the solutions for \(x\) are \(x = \pm \sqrt{1}, \pm \sqrt{16}\)
04

Analyzing the Sign of the Derivative

To determine whether these points are maxima, minima, or neither, we determine the sign of the derivative in the intervals. Choosing representative points for each interval and calculating the sign of the derivative, we find that the function is decreasing on \((- \infty, -4)\), increasing on \((-4, -1)\), decreasing on \((-1, 1)\), increasing on \((1, 4)\), and decreasing on \((4, \infty)\). Therefore, the function has relative maximum at \(x=-4, 1\) and relative minimum at \(x=-1, 4\)

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Most popular questions from this chapter

Analyzing a Function \(\quad\) Let \(f(x)=\frac{\ln x}{x}\) (a) Graph \(f\) on \((0, \infty)\) and show that \(f\) is strictly decreasing on \((e, \infty) .\) (b) Show that if \(e \leq AB^{A}\) . (c) Use part (b) to show that \(e^{\pi}>\pi^{e}\) .

Use implicit differentiation to find an equation of the tangent line to the graph of the equation at the given point. \(\arcsin x+\arcsin y=\frac{\pi}{2}, \quad\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\)

Analyze and sketch a graph of the function. Identify any relative extrema, points of inflection, and asymptotes. Use a graphing utility to verify your results. \(f(x)=\arccos \frac{x}{4}\)

Find the derivative of the function. \(y=25 \arcsin \frac{x}{5}-x \sqrt{25-x^{2}}\)

Using the Area of a Region Find the value of \(a\) such that the area bounded by \(y=e^{-x},\) the \(x\) -axis, \(x=-a,\) and \(x=a\) is \(\frac{8}{3} .\)
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