Question

The minimum value of $2x^2+x-1$ is

• A. $-\frac{1}{4}$
• B. $\frac{3}{4}$
• C. $-\frac{9}{8}$
• D. $\frac{9}{4}$

Answer 1

The correct option is C $-\frac{9}{8}$

$y=2x^2+x-1$
Differentiating with respect to x gives us
$\frac{dy}{dx}=4x+1$
$y^{\prime }=0$ implies
$4x+1=0$
$x=\frac{-1}{4}$.
Hence
$f\left(\frac{-1}{4}\right)=2\left(\frac{1}{16}\right)-\frac{1}{4}-1$
$=\frac{1}{8}-\frac{2}{8}-\frac{8}{8}$
$=\frac{-9}{8}$. ...(i)
Now
$f^{\prime \prime }\left(x\right)=2$
Since
$f^{\prime \prime }\left(\frac{-1}{4}\right)>0$ hence $x=\frac{-1}{4}$ is a point of minimum.
Thus f(x) attains the minimum value at $x=\frac{-1}{4}$ the minimum value being $f\left(\frac{-1}{4}\right)=\frac{-9}{8}$