Show that $f$ is a decreasing function It's given that $f(x)=\frac{1}{x^3}-x^3$ for $x>0$ show that $f$ is a decreasing function. 
My attempt $f'(x)=-3x^{-4}-3x^2$
$x^6=-1$
How to continue by my attempt ?
 A: you need to show $f'$ is always negative.
suppose $f'$ is positive. then $-3x^{-4}-3x^2=k>0$
$-3-3x^6=kx^4$
$3x^6+kx^4+3=0$
since $x^2$ is always greater than or equal to zero, so it $3(x^2)^3+k(x^2)^2$, so if you add $3$ the result is always positive. thus no such $k$ exists
A: Notice, we have $$f(x)=\frac{1}{x^3}-x^3$$
$$\frac{d}{dx}(f(x))=\frac{d}{dx}\left(\frac{1}{x^3}-x^3\right)$$
$$f'(x)=\frac{-3}{x^4}-3x^2$$
$$f'(x)=-3\left(\frac{1}{x^4}+x^2\right)$$
$$f'(x)=-3\left(\left(\frac{1}{x^2}\right)^2+x^2\right)\tag 1$$
We know that $$\left(\frac{1}{x^2}\right)^2> 0\ \ \ \ \forall \ x>0 \tag 2$$ & $$\left(x\right)^2> 0\ \ \ \ \forall \ x>0 \tag 3$$ From (2) & (3) we get $$\left(\frac{1}{x^2}\right)^2+x^2> 0$$$$\iff -3\left(\left(\frac{1}{x^2}\right)^2+x^2\right)<0\ \ \ \ \forall \  \ x>0\tag 4$$ Hence, from (1) & (4), we find that $$f'(x)<0\ \ \ \forall \ \ \ x>0$$  
Hence, the function $f(x)  $ is decreasing for all $x>0$
A: Simply for every $x>0$ we have $x^{-4}>0$ and$x^2>0 $ so $-\frac{3}{x^4}<0$ and$\frac{-3}{x^2}<0$ then by adding this tow term you have $f'(x)<0 $ for all $x>0$
A: From your work it is clear that there is no $x$ such that $f'(x)=0$. Then for $x>0$ note that $f'(x)$ is continuous, hence it is impossible to have $a,b>0$ such that $f'(a)>0, f'(b)<0$ due to intermediate value theorem. Then plug in $x=1$ we know that $f'(1)<0$, hence for all $x>0$ $f'(x)<0$. Hence $f$ is decreasing.
