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

given function , f(x) = x² + ax + 1
differentiate with respect to x,
f'(x) =2x + a

function is strictly increasing on (1, 2)
so, f'(x) = 2x + a > 0 on (1, 2)
e.g., x > -a/2
Therefore, we have to find the least value of a such that
⇒ x > -a/2 when x belongs to (1,2)
⇒ x > -a/2 when 1 < x < 2
here it is clear that we can get least value of a only when we put least value of x e.g., 1
Therefore, the least value of a for function to be increasing on (1,2) is given by
-a/2 = 1
⇒ a = -2

Therefore, the least value of a is -2.

Answer 2

Answer:

-2 is the answer of question