The least value of k for which the function f | vedclass

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(D) Let $f(x) = {x^2} + kx + 1$. For the function to be increasing on the interval $[1, 2]$,its derivative must satisfy $f'(x) \geq 0$ for all $x \in

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$-2$

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(D) Let $f(x) = {x^2} + kx + 1$. For the function to be increasing on the interval $[1, 2]$,its derivative must satisfy $f'(x) \geq 0$ for all $x \in [1, 2]$. Calculating the derivative,we get $f'(x) = 2x + k$. Setting the condition $f'(x) \geq 0$,we have $2x + k \geq 0$,which implies $k \geq -2x$. Since this must hold for all $x$ in the interval $[1, 2]$,$k$ must be greater than or equal to the maximum value of $-2x$ on this interval. For $x \in [1, 2]$,the function $g(x) = -2x$ is a decreasing function. Therefore,the maximum value of $g(x)$ occurs at the lower bound $x = 1$. $g(1) = -2(1) = -2$. Thus,$k \geq -2$. The least value of $k$ is $-2$.

The least value of $k$ for which the function $f(x) = {x^2} + kx + 1$ is an increasing function in the interval $1 \leq x \leq 2$ is

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