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Let f(x) ={underset(x^(2) +xb " " x g...

Let f(x) `={underset(x^(2) +xb " " x ge1)(3-x " "0le x lt1).` Find the set of values of b such that f(x) has a local minima at x=1.

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To solve the problem, we need to find the set of values of \( b \) such that the function \( f(x) \) has a local minimum at \( x = 1 \). The function is defined as follows: \[ f(x) = \begin{cases} 3 - x & \text{for } 0 \leq x < 1 \\ x^2 + bx & \text{for } x \geq 1 \end{cases} \] ### Step 1: Find the first derivative \( f'(x) \) We differentiate \( f(x) \) piecewise: - For \( 0 \leq x < 1 \): \[ f'(x) = \frac{d}{dx}(3 - x) = -1 \] - For \( x \geq 1 \): \[ f'(x) = \frac{d}{dx}(x^2 + bx) = 2x + b \] ### Step 2: Evaluate the first derivative at \( x = 1 \) To find the critical point at \( x = 1 \), we need to set \( f'(1) = 0 \): \[ f'(1) = 2(1) + b = 2 + b \] Setting this equal to zero gives: \[ 2 + b = 0 \implies b = -2 \] ### Step 3: Find the second derivative \( f''(x) \) Next, we find the second derivative \( f''(x) \): - For \( 0 \leq x < 1 \): \[ f''(x) = 0 \] - For \( x \geq 1 \): \[ f''(x) = \frac{d}{dx}(2x + b) = 2 \] ### Step 4: Check the second derivative at \( x = 1 \) Since \( f''(x) = 2 \) for \( x \geq 1 \), we evaluate it at \( x = 1 \): \[ f''(1) = 2 \] ### Step 5: Determine the nature of the critical point Since \( f''(1) > 0 \), this indicates that the function \( f(x) \) has a local minimum at \( x = 1 \). ### Conclusion The value of \( b \) for which \( f(x) \) has a local minimum at \( x = 1 \) is: \[ \boxed{-2} \]
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