Least possible value of $k$ for which $f(x)=x^2+kx+1$ is increasing on $[1,2]$

Question

Find the least possible value of $k$ for which $f(x)=x^2+kx+1$ is increasing on $[1,2]$

$$ f(x)=x^2+kx+1\\ f'(x)=2x+k\geq0\implies \boxed{k\geq-2x}\\ 1\leq x\leq2\implies -2\leq-x\leq-1\implies-4\leq-2x\leq-2\implies k>-4\\ f(2)>f(1)\implies k>-3 $$

The solution given in my reference is $x=-2$, how do I obtain it mathematically ?

Answer 1

$$f(x)=x^2+kx+1$$ $$f'(x)=2x+k\geq0\implies \boxed{k\geq-2x}$$

It's good up to here, but since we must have $k \geq -2x$ for $x \in [1, 2]$ it suffices to have $k = -2$.

Mathematically, if we have $$k = -2$$ and $$1 \leq x \leq 2$$

Then, $$1 \leq x$$ $$-2x \leq -2$$ $$-2x \leq k$$ $$2x + k \geq 0$$

To prove that $-2$ is the lowest possible value for $k$, suppose $$k < -2$$ and let $$x = 1$$ Then, $$2x + k < 0$$ which is obviously a contradiction.

Answer 2

Rearranging the inequality might help:

$$ k\geq -2x\iff -k/2\leq x $$ If $1\leq x\leq 2$, it's clear one must take $-k/2\leq 1$, from which the answer follows.

Answer 3

I think the answer is clear. Because by the condition $k》-2x$ the maximum value of $-2x$ is the least value of k. The maximum value of $-2,x$ is $-2$.