Algebra

Find the minimum value of \frac{x^2}{x - 1 + x^3} for x > 1    

                                                  (x²) / (x-1+x³)   

At x = 1, it equals 1 / 1.  As x gets larger than 1, the denominator increases faster than the numerator.    

That's a condition that causes the fraction to get smaller and smaller, but it never quite reaches zero.    

So, the fraction approaches zero.  It approaches zero but never gets there.  It's called an asymptote.    

I think they call this approaching zero asymptotically.    

_.    

\(f(x)= \dfrac{x^2}{x - 1 + x^3}\\\\ u=x^2;\\v= x-1+x^3\\ f'(x)=\dfrac{u'v-uv'}{v^2}=\dfrac{2x\cdot (x-1+x^3)-x^2\cdot(1+3x^2)}{x-1+x^3}=0\\ \)

\(2x\cdot (x-1+x^3)-x^2\cdot(1+3x^2)=0\\ x\in \{0,-1.5214\}\\ f_1(x)= \dfrac{x^2}{x - 1 + x^3}= \dfrac{0^2}{0-1+0^3}=\color{blue}0\ max.\\ f_2(x)= \dfrac{x^2}{x - 1 + x^3}=\dfrac{(-1.5214)^2}{-1.5214-1+(-1.5214)^3}=\color{blue}-0.3830363\ min.\\ \color{blue}WolframAlpha\)

These are the values ​​for x<0.5.

The values ​​for x>1 are very well described in answer 1.

 !