Let f(x) = begin{cases} 3 - x, & 0 le x

Question

(B) For $f(x)$ to have a local minimum at $x = 1$,the function must satisfy $f(1) \le f(1 - h)$ and $f(1) \le f(1 + h)$ for small $h > 0$.First,we eva

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$(0, e]$

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(B) For $f(x)$ to have a local minimum at $x = 1$,the function must satisfy $f(1) \le f(1 - h)$ and $f(1) \le f(1 + h)$ for small $h > 0$.First,we evaluate the left-hand limit at $x = 1$:$f(1^-) = \lim_{x 	o 1^-} (3 - x) = 3 - 1 = 2$.Next,we evaluate the function value at $x = 1$:$f(1) = 1^2 + \log_e b = 1 + \log_e b$.For $x > 1$,$f(x) = x^2 + \log_e b$. Since $f'(x) = 2x > 0$ for $x > 1$,the function is strictly increasing for $x > 1$. Thus,$f(x) \ge f(1)$ for all $x > 1$ in the neighborhood of $1$.For $x  f(1^-) = 2$ for all $x For $x = 1$ to be a local minimum,we require $f(1) \le f(x)$ for all $x$ in a neighborhood of $1$. Specifically,we need $f(1) \le f(1^-)$:$1 + \log_e b \le 2$$\log_e b \le 1$$b \le e^1 = e$.Since the logarithm $\log_e b$ is defined only for $b > 0$,the set of values for $b$ is $(0, e]$.

Let $f(x) = \begin{cases} 3 - x, & 0 \le x < 1 \\ x^2 + \log_e b, & x \ge 1 \end{cases}$. The set of values of $b$ such that $f(x)$ has a local minimum at $x = 1$ is

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