If R is the least value of a such that the fu | vedclass

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(A) Given the function $f(x) = x^{2} + ax + 1$.The derivative is $f'(x) = 2x + a$.For $f(x)$ to be increasing on $[1, 2]$,we must have $f'(x) \geq 0$

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$2$

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(A) Given the function $f(x) = x^{2} + ax + 1$.The derivative is $f'(x) = 2x + a$.For $f(x)$ to be increasing on $[1, 2]$,we must have $f'(x) \geq 0$ for all $x \in [1, 2]$.$2x + a \geq 0 \implies a \geq -2x$ for all $x \in [1, 2]$.The minimum value of $-2x$ on $[1, 2]$ occurs at $x = 2$,which is $-2(2) = -4$.Thus,$R = -4$.For $f(x)$ to be decreasing on $[1, 2]$,we must have $f'(x) \leq 0$ for all $x \in [1, 2]$.$2x + a \leq 0 \implies a \leq -2x$ for all $x \in [1, 2]$.The maximum value of $a$ is determined by the minimum value of $-2x$ on the interval,but since $a$ must be less than or equal to $-2x$ for all $x$,$a$ must be less than or equal to the minimum value of $-2x$ on $[1, 2]$.The minimum value of $-2x$ on $[1, 2]$ is $-4$. Wait,let us re-evaluate: $a \leq -2x$ for all $x \in [1, 2]$. The condition $a \leq -2x$ must hold for the entire interval,so $a \leq \min(-2x) = -4$. Thus,the greatest value $S = -4$.Let us re-check the increasing condition: $a \geq -2x$ for all $x \in [1, 2]$. This means $a \geq \max(-2x) = -2(1) = -2$. So $R = -2$.Then $|R - S| = |-2 - (-4)| = |-2 + 4| = 2$.

If $R$ is the least value of $a$ such that the function $f(x) = x^{2} + ax + 1$ is increasing on $[1, 2]$ and $S$ is the greatest value of $a$ such that the function $f(x) = x^{2} + ax + 1$ is decreasing on $[1, 2]$,then the value of $|R - S|$ is ..... .

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