We are given the following conditions: - \( f(0) = 2 \) - \( f'(4) = -\frac{3}{4} \) - The chord \( AB \) is parallel to the tangent at \( (4, f(4)) \).
Step 1: Slope of the tangent at \( x = 4 \). The slope of the tangent at \( x = 4 \) is given by \( f'(4) \), which is \( -\frac{3}{4} \).
Step 2: Slope of the chord \( AB \). The slope of the chord \( AB \) is given by the difference in the \( y \)-coordinates of \( A \) and \( B \) divided by the difference in the \( x \)-coordinates of \( A \) and \( B \): \[ \text{slope of } AB = \frac{\beta - \alpha}{8 - 0} = \frac{\beta - \alpha}{8} \] Since the chord \( AB \) is parallel to the tangent at \( (4, f(4)) \), the slope of the chord is equal to the slope of the tangent, i.e., \[ \frac{\beta - \alpha}{8} = -\frac{3}{4} \]
Step 3: Substitute \( \alpha = f(0) = 2 \). Substituting \( \alpha = 2 \) into the equation: \[ \frac{\beta - 2}{8} = -\frac{3}{4} \] Step 4: Solve for \( \beta \). Multiply both sides by 8 to eliminate the denominator: \[ \beta - 2 = -6 \] Finally, add 2 to both sides: \[ \beta = -4 \] Thus, the value of \( \beta \) is \( -4 \).
If \( f(x) = \begin{cases} 2x - 3, & -3 \leq x \leq -2 \\x + 1, & -2<x \leq 0 \end{cases} \), check the differentiability of \( f(x) \) at \( x = -2 \).