If f(x) = x^2 + kx + 1 is an increasing funct | vedclass

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(C) For a function $f(x)$ to be increasing on an interval,its derivative $f'(x)$ must be greater than or equal to $0$ on that interval.Given $f(x) = x

Vedclass · Accepted Answer

$-2$

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(C) For a function $f(x)$ to be increasing on an interval,its derivative $f'(x)$ must be greater than or equal to $0$ on that interval.Given $f(x) = x^2 + kx + 1$,we find the derivative:$f'(x) = 2x + k$For $f(x)$ to be increasing on $[1, 2]$,we require $f'(x) \geq 0$ for all $x \in [1, 2]$.$2x + k \geq 0$$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.Let $g(x) = -2x$. On the interval $[1, 2]$,the function $g(x)$ is decreasing.Therefore,the maximum value of $g(x)$ occurs at the left endpoint $x = 1$.$g(1) = -2(1) = -2$.Thus,$k \geq -2$.The minimum value of $k$ is $-2$.

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

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