Question

For real $x$ find the minimum value of $4x^2-2x+1$

Answer 1

Let, $f\left(x\right)=4x^2-2x+1$.

Now $f^{\prime }\left(x\right)=8x-2$

And $f^{\prime \prime }\left(x\right)=8$.

For maximum of minimum value of $f\left(x\right)$ we must have $f^{\prime }\left(x\right)=0$.

Then we get,

$x=\frac{1}{4}$.

For $x=\frac{1}{4}$ we have $f^{\prime \prime }\left(x\right)>0$.

So $x=\frac{1}{4}$ gives a minimum value for $f\left(x\right)$.

So the minimum value of $f\left(x\right)$ is

$4.\frac{1}{16}-2.\frac{1}{4}+1$

$=\frac{1}{4}-\frac{1}{2}+1$

$=\frac{3}{4}$.