Range of the function f(x) = cot^{-1} left( l | vedclass

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(B) Let $g(x) = 5x^2 - 8x + 4$.Completing the square,$g(x) = 5(x^2 - \frac{8}{5}x) + 4 = 5(x - \frac{4}{5})^2 + 4 - 5(\frac{16}{25}) = 5(x - \frac{4}{

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$[\frac{\pi}{4}, \pi)$

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(B) Let $g(x) = 5x^2 - 8x + 4$.Completing the square,$g(x) = 5(x^2 - \frac{8}{5}x) + 4 = 5(x - \frac{4}{5})^2 + 4 - 5(\frac{16}{25}) = 5(x - \frac{4}{5})^2 + \frac{4}{5}$.Since $5(x - \frac{4}{5})^2 \geq 0$,the range of $g(x)$ is $[\frac{4}{5}, \infty)$.Let $u = \log_{4/5}(g(x))$. Since the base $4/5 For $t \in [\frac{4}{5}, \infty)$,$\log_{4/5}(t) \in (\log_{4/5}(\infty), \log_{4/5}(\frac{4}{5})] = (-\infty, 1]$.Now,$f(x) = \cot^{-1}(u)$ where $u \in (-\infty, 1]$.Since $\cot^{-1}(u)$ is a strictly decreasing function,the range is $[\cot^{-1}(1), \cot^{-1}(-\infty)) = [\frac{\pi}{4}, \pi)$.

Range of the function $f(x) = \cot^{-1} \left( \log_{4/5} (5x^2 - 8x + 4) \right)$ is:

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