Question

Find the least value of a such that the function f given by $f\left(x\right)=x^2+ax+1$ is strictly increasing on (1,2).

Answer 1

Given, $f\left(x\right)=x^2+ax+1\Rightarrow f^{\prime }\left(x\right)=2x+a$
In interval (1,2), $1<x<2\Rightarrow 2<2x<4\Rightarrow (2+a)<(2x+a)<(4+a)$
Since, f(x) is strictly increasing function, then $(2+a)>0$ [For this $f^{\prime }\left(x\right)>0$ and $(2x+a)>(2+a)$]
$\therefore (2+a)>0\Rightarrow a>-2$. Hence, the least value of a=-2