The function f(x) = (x - 3)^2 satisfies all t | vedclass

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(B) According to the Mean Value Theorem,there exists a point $c \in (3, 4)$ such that $f'(c) = \frac{f(4) - f(3)}{4 - 3}$.First,calculate the slope of

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$\left( \frac{7}{2}, \frac{1}{4} \right)$

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(B) According to the Mean Value Theorem,there exists a point $c \in (3, 4)$ such that $f'(c) = \frac{f(4) - f(3)}{4 - 3}$.First,calculate the slope of the chord joining $(3, 0)$ and $(4, 1)$:$m = \frac{1 - 0}{4 - 3} = 1$.Now,find the derivative of $f(x) = (x - 3)^2$:$f'(x) = 2(x - 3)$.Set the derivative equal to the slope of the chord:$2(c - 3) = 1 \Rightarrow c - 3 = \frac{1}{2} \Rightarrow c = \frac{7}{2}$.Now,find the corresponding $y$-coordinate by substituting $x = \frac{7}{2}$ into the function:$y = \left( \frac{7}{2} - 3 \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.Thus,the point is $\left( \frac{7}{2}, \frac{1}{4} \right)$.

The function $f(x) = (x - 3)^2$ satisfies all the conditions of the Mean Value Theorem in $[3, 4]$. $A$ point on $y = (x - 3)^2$,where the tangent is parallel to the chord joining $(3, 0)$ and $(4, 1)$,is

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