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Find all critical points of the function $\newline$ $f(x)=\cos^{-1}(x)+3x$ for $-1 < x < 1.$

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Find derivatives of logarithmic functions

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Q. Find all critical points of the function

\newline

f(x)=\cos^{-1}(x)+3x

for

-1 < x < 1.

Find Derivative and Set Equal: To find the critical points of the function $f(x) = \cos^{-1}(x) + 3x$ , we need to find the derivative of the function and set it equal to $0$ .
Calculate Derivative: The derivative of $f(x)$ with respect to $x$ is given by $f'(x) = \frac{d}{dx}[\cos^{-1}(x)] + \frac{d}{dx}[3x]$ .
Set Derivative Equal to Zero: Using the chain rule and the fact that the derivative of $\cos^{-1}(x)$ is $-\frac{1}{\sqrt{1-x^2}}$ , we get $f'(x) = -\frac{1}{\sqrt{1-x^2}} + 3$ .
Solve for Critical Points: To find the critical points, we set the derivative equal to zero: $-\frac{1}{\sqrt{1-x^2}} + 3 = 0$ .
Add $1$ to Both Sides: Solving for $x$ , we multiply both sides by $\sqrt{1-x^2}$ to get $-1 + 3\sqrt{1-x^2} = 0$ .
Divide by $3$ : Next, we add $1$ to both sides to obtain $3\sqrt{1-x^2} = 1$ .
Square Both Sides: Dividing both sides by $3$ gives us $\sqrt{1-x^2} = \frac{1}{3}$ .
Simplify Equation: Squaring both sides to eliminate the square root gives us $1 - x^2 = \left(\frac{1}{3}\right)^2$ .
Find $x$ Values: This simplifies to $1 - x^2 = \frac{1}{9}$ .
Consider Interval: Subtracting $1$ from both sides yields $-x^2 = \frac{1}{9} - 1$ .
Final Critical Points: Simplifying the right side, we get $-x^2 = -\frac{8}{9}$ .
Final Critical Points: Simplifying the right side, we get $-x^2 = -\frac{8}{9}$ . Dividing by $-1$ , we find $x^2 = \frac{8}{9}$ .
Final Critical Points: Simplifying the right side, we get $-x^2 = -\frac{8}{9}$ .Dividing by $-1$ , we find $x^2 = \frac{8}{9}$ .Taking the square root of both sides, we get $x = \pm\sqrt{\frac{8}{9}}$ .
Final Critical Points: Simplifying the right side, we get $-x^2 = -\frac{8}{9}$ .Dividing by $-1$ , we find $x^2 = \frac{8}{9}$ .Taking the square root of both sides, we get $x = \pm\sqrt{\frac{8}{9}}$ .Simplifying the square root, we find $x = \pm\frac{\sqrt{8}}{3}$ .
Final Critical Points: Simplifying the right side, we get $-x^2 = -\frac{8}{9}$ .Dividing by $-1$ , we find $x^2 = \frac{8}{9}$ .Taking the square root of both sides, we get $x = \pm\sqrt{\frac{8}{9}}$ .Simplifying the square root, we find $x = \pm\frac{\sqrt{8}}{3}$ .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 $x = \pm\frac{\sqrt{8}}{3}$ are approximately $\pm0.9428$ , which are within the interval.
Final Critical Points: Simplifying the right side, we get $-x^2 = -\frac{8}{9}$ . Dividing by $-1$ , we find $x^2 = \frac{8}{9}$ . Taking the square root of both sides, we get $x = \pm\sqrt{\frac{8}{9}}$ . Simplifying the square root, we find $x = \pm\frac{\sqrt{8}}{3}$ . 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 $x = \pm\frac{\sqrt{8}}{3}$ are approximately $\pm0.9428$ , which are within the interval. Therefore, the critical points of the function $f(x) = \cos^{-1}(x) + 3x$ for $-1 < x < 1$ are $-1$ $0$ and $-1$ $1$ .

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