Question

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

Answer 1

The given function is $f\left(x\right)=\frac{2x^2-1}{x^4},x>0$.

$f\left(x\right)=\frac{2x^2-1}{x^4}$

$\Rightarrow f\left(x\right)=\frac{2}{x^2}-\frac{1}{x^4}$

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=-\frac{4}{x^3}+\frac{4}{x^5}$

$\Rightarrow f\text{'}\left(x\right)=-4\left(\frac{x^2-1}{x^5}\right)$

$\Rightarrow f\text{'}\left(x\right)=\frac{-4\left(x+1\right)\left(x-1\right)}{x^5}$

For f(x) to be decreasing,

$f\text{'}\left(x\right)<0$

$\Rightarrow \frac{-4\left(x+1\right)\left(x-1\right)}{x^5}<0$

$\Rightarrow \left(x+1\right)\left(x-1\right)>0\left(x^5>0\mathrm{for}x>0\right)$

$\Rightarrow \left[x-\left(-1\right)\right]\left(x-1\right)>0$

$\Rightarrow x\in \left(-\infty ,-1\right)\cup \left(1,\infty \right)$ [For a < b, if (x − a)(x − b) > 0 ⇒ x ∈ (−∞, a) ∪ (b, ∞)]

But, x > 0 (Given)

∴ x ∈ (1, ∞)

Thus, the given function f(x) decreases in the interval (1, ∞).


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