f(x) = cos^{-1}(2x^2 - 1) is not differentiab | vedclass

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(C) The function is given by $f(x) = \cos^{-1}(2x^2 - 1)$.We know that $\cos^{-1}(\cos 	heta) = 	heta$ for $	heta \in [0, \pi]$.Let $x = \cos 	het

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(C) The function is given by $f(x) = \cos^{-1}(2x^2 - 1)$.We know that $\cos^{-1}(\cos 	heta) = 	heta$ for $	heta \in [0, \pi]$.Let $x = \cos 	heta$,then $2x^2 - 1 = 2\cos^2 	heta - 1 = \cos(2	heta)$.Thus,$f(x) = \cos^{-1}(\cos(2	heta)) = 2	heta = 2\cos^{-1}(x)$ for $x \in [0, 1]$.For $x \in [-1, 0]$,let $x = -\cos 	heta = \cos(\pi - 	heta)$.Then $2x^2 - 1 = \cos(2(\pi - 	heta)) = \cos(2\pi - 2	heta) = \cos(2	heta)$.So $f(x) = 2\cos^{-1}(x)$ is not the correct simplification for the whole domain.Actually,$f(x) = 2\cos^{-1}|x|$.The function $f(x) = 2\cos^{-1}(x)$ for $x \ge 0$ and $f(x) = 2\cos^{-1}(-x)$ for $x The derivative of $\cos^{-1}(x)$ is $\frac{-1}{\sqrt{1-x^2}}$.The function $f(x)$ is not differentiable where the argument of the inverse cosine function reaches its boundaries,i.e.,$2x^2 - 1 = 1$ or $2x^2 - 1 = -1$.$2x^2 = 2 \implies x^2 = 1 \implies x = \pm 1$.Also,the function $f(x) = 2\cos^{-1}|x|$ has a cusp at $x = 0$ because the derivative from the left and right are not equal.At $x = 0$,$f(x) = 2\cos^{-1}(0) = \pi$.Left derivative: $\frac{d}{dx}(2\cos^{-1}(-x)) = 2 \cdot \frac{-1}{\sqrt{1-(-x)^2}} \cdot (-1) = \frac{2}{\sqrt{1-x^2}}$. At $x=0$,this is $2$.Right derivative: $\frac{d}{dx}(2\cos^{-1}(x)) = \frac{-2}{\sqrt{1-x^2}}$. At $x=0$,this is $-2$.Since $2 
eq -2$,the function is not differentiable at $x = 0$.

$f(x) = \cos^{-1}(2x^2 - 1)$ is not differentiable at $x = a$,then $a$ equals:

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