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The function f(x)=\frac{2x^2-1}{x^4},x>0decreases in the interval _______.

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Given f(x)=\frac{2x^2-1}{x^4},x>0

After applying the derivative, we get

$$ \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\frac{2 \mathrm{x}^{2}-1}{\mathrm{x}^{4}}\right)}{\mathrm{d} \mathrm{x}}
Apply the quotient rule and 0 is the differentiation of the constant term, so

\\ f^{\prime}(x)=\frac{x^{4} \cdot \frac{d\left(2 x^{2}-1\right)}{d x}-\left(2 x^{2}-1\right) \cdot \frac{d\left(x^{4}\right)}{d x}}{\left(x^{4}\right)^{2}} \\ f^{\prime}(x)=\frac{1}{x^{8}}\left(x^{4} \cdot \frac{d\left(2 x^{2}-1\right)}{d x}-\left(2 x^{2}-1\right) \cdot \frac{d\left(x^{4}\right)}{d x}\right) \\ f^{\prime}(x)=\frac{1}{x^{8}}\left(x^{4} \cdot(4 x)-\left(2 x^{2}-1\right) \cdot\left(4 x^{3}\right)\right) \\ f^{\prime}(x)=\frac{1}{x^{8}}\left(4 x^{5}-\left(2 x^{2}\right) \cdot\left(4 x^{3}\right)-\left(4 x^{3}\right)\right) \\ f^{\prime}(x)=\frac{1}{x^{8}}\left(4 x^{5}-8 x^{5}-\left(4 x^{3}\right)\right) \\ f^{\prime}(x)=\frac{x^{3}}{x^{8}}\left(4 x^{2}-8 x^{2}-4\right) \\ f^{\prime}(x)=\frac{1}{x^{5}}\left(-4 x^{2}-4\right) \\ f^{\prime}(x)=\frac{-4}{x^{5}}\left(x^{2}-1\right)
Equate this with 0 and get
\\ f^{\prime}(x)=0$ \\$\Rightarrow \frac{-4}{x^{5}}\left(x^{2}-1\right)=0$ \\$\Rightarrow x^{2}-1=0$ \\$\Rightarrow x^{2}=1$ \\$\Rightarrow x=\pm 1
(-\infty,-1),(-1,0),(0,1)and (1, \infty) are the intervals formed by these two critical numbers
(i) in the interval (-\infty,-1), f^{\prime}(x)>0
\therefore f(x)$ is increasing in $(-\infty,-1)

(ii) in the interval (-1,0), \mathrm{f}^{\prime}(\mathrm{x}) \leq 0
\therefore \mathrm{f}(\mathrm{x}) is decreasing in (-1,0)
(iii) in the interval (0,1), \mathrm{f'}(x)>0
\therefore \mathrm{f}(\mathrm{x}) is increasing in (0,1)
(iii) in the interval (1, \infty), f^{\prime}(x)<0
\therefore \mathrm{f}(\mathrm{x}) is decreasing in (1, \infty)
f(x)=\frac{2 x^{2}-1}{x^{4}}, x>0

Therefore, the function f(x)=\frac{2 x^{2}-1}{x^{4}}, x>0decreases in the interval (1, \infty).

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