Question

The number of values of $x$ for which the function $f\left(x\right)=x^2-3x+2$ is concave down, is equal to

Answer 1

$f\left(x\right)=x^2-3x+2$
$f^{\prime }\left(x\right)=2x-3$
$f^{\prime \prime }\left(x\right)=2>0\forall x\in R$
$\therefore f\left(x\right)$ is concave up $\forall x\in R$
Hence, there exists no such value of $x$ for which the function is concave down.