For the function f(x) = x^2 - 6x + 8 where 2 | vedclass

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(C) Given the function $f(x) = x^2 - 6x + 8$. To find the value of $x$ for which $f'(x)$ vanishes,we first find the derivative of $f(x)$ with respect

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$3$

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(C) Given the function $f(x) = x^2 - 6x + 8$. To find the value of $x$ for which $f'(x)$ vanishes,we first find the derivative of $f(x)$ with respect to $x$: $f'(x) = \frac{d}{dx}(x^2 - 6x + 8) = 2x - 6$. Now,set the derivative equal to zero: $2x - 6 = 0$. Solving for $x$: $2x = 6 \Rightarrow x = 3$. Since $x = 3$ lies within the interval $[2, 4]$,the value of $x$ for which $f'(x)$ vanishes is $3$.

For the function $f(x) = x^2 - 6x + 8$ where $2 \le x \le 4$,the value of $x$ for which $f'(x)$ vanishes is:

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