Question

For what value of $x$, $8x^2-7x+2$ has the minimum value?

Answer 1

Let $f\left(x\right)=8x^2-7x+2$

Then $f^{\prime }\left(x\right)=16x=-7=0$

$\Rightarrow 16x=7$

$\Rightarrow x=\frac{7}{16}$

Now $f^{\prime \prime }\left(x\right)=16$ which is positive.

Therefore, we get minimum value at $x=\frac{7}{16}$.