If f(x) = x^2 + kx + 1 is a strictly increasi | vedclass

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

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$-2$

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(D) Given the function $f(x) = x^2 + kx + 1$.For $f(x)$ to be strictly increasing on $[1, 2]$,its derivative must satisfy $f'(x) > 0$ for all $x \in [1, 2]$.Calculating the derivative: $f'(x) = 2x + k$.Setting the condition: $2x + k > 0$ for all $x \in [1, 2]$.This implies $k > -2x$ for all $x \in [1, 2]$.To satisfy this for all $x$ in the interval,$k$ must be greater than the maximum value of $-2x$ on $[1, 2]$.The function $g(x) = -2x$ is a decreasing function,so its maximum value on $[1, 2]$ occurs at $x = 1$.Thus,$g(1) = -2(1) = -2$.Therefore,$k > -2$. The infimum (minimum value) of $k$ is $-2$.

If $f(x) = x^2 + kx + 1$ is a strictly increasing function on the interval $[1, 2]$,then what is the minimum value of $k$?

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