Question

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

• A. $\frac{1}{4}$
• B. $\frac{3}{2}$
• C. $-\frac{9}{8}$
• D. $\frac{9}{4}$

Answer 1

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

$f\left(x\right)=2x^2+x-1$
Differentiating the above function with respect to x.
$f^{\prime }\left(x\right)=4x+1$
$=0$
Or
$4x+1=0$
$x=\frac{-1}{4}$.
Now
$f^{\prime \prime }\left(x\right)=4$
or
$f^{\prime \prime }(x{)}_{x=\frac{-1}{4}}>0$
Hence the function attains a minimum value at x equal to -0.25.
Therefore
$f\left(\frac{-1}{4}\right)=2\left(\frac{1}{16}\right)-\frac{1}{4}-1$
$=\frac{1}{8}-\frac{1}{4}-1$
$=\frac{1}{8}-\frac{2}{8}-\frac{8}{8}$
$=\frac{-9}{8}$