Is there any way to find minimum without the use of derivatve? The function is: $$\sqrt{(x+1)^2+\left(2x^2-\frac{1}{4}\right)^2}$$
It simplifies to: $$\sqrt{4x^4+2x+\frac{17}{16}}$$
 A: Geometric Approach
See this as $$\sqrt{(x-(-1))^2+(2x^2-1/4)^2}$$
ie. distance of $(x,2x^2)${a point on curve $y=2x^2$ } from $(-1,1/4)$
That would be minimum along the normal to the curve $y=2x^2$ passing through $(-1,1/4)$
Slope of normal through point $(x,2x^2)$ to the curve = $-1/4x$
So, $$-\dfrac1{4x}=\dfrac{2x^2-1/4}{x-(-1)}==>x=-1/2$$
ie. Equating it to the slope of two points $(x,2x^2)$ and $(-1,1/4)$
So, the normal through $(-1,1/4)$ passes through $(-1/2,1/2)$ in the curve $y=2x^2$
Find distance between these points.
A: I like @exploringnet's solution. Another way to do this is by a bisection algorithm. It might take you a while and is not nearly as elegant.
Let's focus on minimizing $f(x)=2x^4+x$ (a strictly increasing transformation of the original function). Its minimum will be a minimum of the original function.
Start with $f(0)=0$. Note that from $0$ the function is strictly increasing in the positive direction. That will do us no good. Let's move in the negative direction.
$f(-2)=32-2=30$. It's bigger. Moving any further away in the negative direction will do us no good. Go half way between $0$ and $-2$ to $-1$.
$f(-1)=1$. Still bigger than 0. Move half way again to $-\frac{1}{2}$
$f(-\frac{1}{2})=\frac{1}{8}-\frac{1}{2}$. Lower now than $0$!
Try $f(-\frac{1}{4})$ and $f(-\frac{3}{4})$. Both are higher than $f(-\frac{1}{2})$
Try $f(-\frac{3}{8})$ and $f(-\frac{5}{8})$. Both are higher than $f(-\frac{1}{2})$
Keep going until you are satisfied $-\frac{1}{2}$ is the minimum. 
