Find Minimum of y: $y=2a+\sqrt{4a^2-8a+3}$

[anemone]

Find the minimum of the function $y=2a+\sqrt{4a^2-8a+3}$.

 

[Albert1]

Find the minimum of the function $y=2a+\sqrt{4a^2-8a+3}$.

Spoiler

for $a<0$
if $a\rightarrow -\infty , \,\, f(a)\rightarrow 2$
if $ a\rightarrow 0^-,\,\,f(a)\rightarrow \sqrt 3$
for $a\geq 0$
by using $AP\geq GP$
if $4a^2=4a^2-8a+3$ then $a=\dfrac {3}{8}$
and $f(\dfrac {3}{8})=1.5$
for: $\dfrac {1}{2}<a<\dfrac {3}{2}$ we get :$y=f(a)$ undefined
$4a^2-8a+3\geq 0$
if :$4a^2-8a+3=0 ,$ then $a=\dfrac {1}{2}$ or $a=\dfrac {3}{2}$
and the minimum of the function =$2\times \dfrac {1}{2}=1\,\,(with \,\, a=\dfrac {1}{2})$

 

[anemone]

Solution of other:

Spoiler

Note that the function of $y$ is concave and continuous over the domain $\left(-\infty,\,\dfrac{1}{2}\right]\cup \left[\dfrac{3}{2},\,\infty\right)$, so it will have its minimum at anyone of the end points $-\infty,\,\dfrac{1}{2},\,\dfrac{3}{2},\,\infty$ and upon checking we get $y_{\text{min}}=2\left(\dfrac{1}{2}\right)+0=1$.