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Find all critical points of the function \newlinef(x)=cos⁡−1(x)+3xf(x)=\cos^{-1}(x)+3x for -1 < x < 1.

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Full solution

Q. Find all critical points of the function \newlinef(x)=cos⁡−1(x)+3xf(x)=\cos^{-1}(x)+3x for −1<x<1.-1 < x < 1.
  1. Find Derivative and Set Equal: To find the critical points of the function f(x)=cos⁡−1(x)+3xf(x) = \cos^{-1}(x) + 3x, we need to find the derivative of the function and set it equal to 00.
  2. Calculate Derivative: The derivative of f(x)f(x) with respect to xx is given by f′(x)=ddx[cos⁡−1(x)]+ddx[3x]f'(x) = \frac{d}{dx}[\cos^{-1}(x)] + \frac{d}{dx}[3x].
  3. Set Derivative Equal to Zero: Using the chain rule and the fact that the derivative of cos⁡−1(x)\cos^{-1}(x) is −11−x2-\frac{1}{\sqrt{1-x^2}}, we get f′(x)=−11−x2+3f'(x) = -\frac{1}{\sqrt{1-x^2}} + 3.
  4. Solve for Critical Points: To find the critical points, we set the derivative equal to zero: −11−x2+3=0-\frac{1}{\sqrt{1-x^2}} + 3 = 0.
  5. Add 11 to Both Sides: Solving for xx, we multiply both sides by 1−x2\sqrt{1-x^2} to get −1+31−x2=0-1 + 3\sqrt{1-x^2} = 0.
  6. Divide by 33: Next, we add 11 to both sides to obtain 31−x2=13\sqrt{1-x^2} = 1.
  7. Square Both Sides: Dividing both sides by 33 gives us 1−x2=13\sqrt{1-x^2} = \frac{1}{3}.
  8. Simplify Equation: Squaring both sides to eliminate the square root gives us 1−x2=(13)21 - x^2 = \left(\frac{1}{3}\right)^2.
  9. Find xx Values: This simplifies to 1−x2=191 - x^2 = \frac{1}{9}.
  10. Consider Interval: Subtracting 11 from both sides yields −x2=19−1-x^2 = \frac{1}{9} - 1.
  11. Final Critical Points: Simplifying the right side, we get −x2=−89-x^2 = -\frac{8}{9}.
  12. Final Critical Points: Simplifying the right side, we get −x2=−89-x^2 = -\frac{8}{9}. Dividing by −1-1, we find x2=89x^2 = \frac{8}{9}.
  13. Final Critical Points: Simplifying the right side, we get −x2=−89-x^2 = -\frac{8}{9}.Dividing by −1-1, we find x2=89x^2 = \frac{8}{9}.Taking the square root of both sides, we get x=±89x = \pm\sqrt{\frac{8}{9}}.
  14. Final Critical Points: Simplifying the right side, we get −x2=−89-x^2 = -\frac{8}{9}.Dividing by −1-1, we find x2=89x^2 = \frac{8}{9}.Taking the square root of both sides, we get x=±89x = \pm\sqrt{\frac{8}{9}}.Simplifying the square root, we find x=±83x = \pm\frac{\sqrt{8}}{3}.
  15. Final Critical Points: Simplifying the right side, we get −x2=−89-x^2 = -\frac{8}{9}.Dividing by −1-1, we find x2=89x^2 = \frac{8}{9}.Taking the square root of both sides, we get x=±89x = \pm\sqrt{\frac{8}{9}}.Simplifying the square root, we find x=±83x = \pm\frac{\sqrt{8}}{3}.Since we are looking for critical points within the interval -1 < x < 1, we only consider the solutions that fall within this range. The solutions x=±83x = \pm\frac{\sqrt{8}}{3} are approximately ±0.9428\pm0.9428, which are within the interval.
  16. Final Critical Points: Simplifying the right side, we get −x2=−89-x^2 = -\frac{8}{9}. Dividing by −1-1, we find x2=89x^2 = \frac{8}{9}. Taking the square root of both sides, we get x=±89x = \pm\sqrt{\frac{8}{9}}. Simplifying the square root, we find x=±83x = \pm\frac{\sqrt{8}}{3}. Since we are looking for critical points within the interval -1 < x < 1, we only consider the solutions that fall within this range. The solutions x=±83x = \pm\frac{\sqrt{8}}{3} are approximately ±0.9428\pm0.9428, which are within the interval. Therefore, the critical points of the function f(x)=cos⁡−1(x)+3xf(x) = \cos^{-1}(x) + 3x for -1 < x < 1 are −1-100 and −1-111.

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