Let f(x) be a differentiable function,A(0, al | vedclass

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(A) Point $A(0, \alpha)$ lies on the curve $y=f(x)$,so $f(0)=\alpha$. Given that $f(0)=2$,we have $\alpha=2$. The slope of the chord $AB$ connecting p

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-$4$

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(A) Point $A(0, \alpha)$ lies on the curve $y=f(x)$,so $f(0)=\alpha$. Given that $f(0)=2$,we have $\alpha=2$. The slope of the chord $AB$ connecting points $A(0, 2)$ and $B(8, \beta)$ is given by: $m_{chord} = \frac{\beta - 2}{8 - 0} = \frac{\beta - 2}{8}$. The slope of the tangent to the curve $y=f(x)$ at $x=4$ is $f^{\prime}(4) = \frac{-3}{4}$. Since the chord $AB$ is parallel to the tangent at $x=4$,their slopes must be equal: $\frac{\beta - 2}{8} = \frac{-3}{4}$ Multiplying both sides by $8$: $\beta - 2 = -3 	imes 2$ $\beta - 2 = -6$ $\beta = -6 + 2 = -4$ Thus,the value of $\beta$ is $-4$.

Let $f(x)$ be a differentiable function,$A(0, \alpha)$ and $B(8, \beta)$ be two points on the curve $y=f(x)$. Given $f(0)=2$ and $f^{\prime}(4)=\frac{-3}{4}$. If the chord $AB$ of the curve is parallel to the tangent drawn at the point $(4, f(4))$,then $\beta=$

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