Question

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

Answer 1

We have,

f'(x) = x2+ax+1

f'(x) = 2x+a

Now, function f will be increasing in (1, 2), if f'(x)>0 in (1, 2).

f'(x)>0

⇒ 2x + a > 0

⇒ 2x > −a

⇒ x>-$\frac a2$

Therefore, we have to find the least value of a such that

x>-$\frac a2$, when x ε (1,2).

x>-$\frac a2$, (when 1<x<2)

Thus, the least value of a for f to be increasing on (1, 2) is given by,

-$\frac a2$ = 1

⇒ a=-2

Hence, the required value of a is −2.