# Find the minimum value without using derivative

Find the minimum value of $$f(x) = {3\over \sqrt{x}+1} - {12\over \sqrt{x}+3}$$

The domain of $$f(x)$$ is $$x ∈ (0,∞)$$. Then, using derivatives, I can find the minimum value is $$f(1)=-1.5$$. However, this uses derivatives.

If $$t=\sqrt{x}\geq 0$$ we get $$f(x) = {3\over \sqrt{x}+1} - {12\over \sqrt{x}+3}$$

$$= {3\over t+1} - {12\over t+3}$$

$$= {3(t+3)-12(t+1)\over (t+1)(t+3)}$$

$$= {-3(3t+1)\over (t+1)(t+3)}$$

Now try to find such real $$m$$ that quadratic equation $${-3(3t+1)\over (t+1)(t+3)} =m$$ i.e. $$mt^2+(4m+9)t +3m+3=0$$ has exactly one solution, so the discriminant $$4m^2+60m+81=0$$

and we get $$m_1 = -{3\over 2}$$ (at $$t=1$$ ) and $$m_2 = -{27\over 2}$$ (at $$t =-{27\over 2}<0$$) so this on can not be. So the minimum value of this expression is $$\boxed{-{3\over 2}}$$

Hint: Start with $$(\sqrt{x}-1)^2\geq 0$$ then we get $$3+3x\geq 6\sqrt{x}$$ and then $$18\sqrt{x}+27+3x\geq 24x+24$$