Key Concepts and Formulas

• Maximum Value of a Function at a Point: For a function $f(x)$ to have a maximum value at $x=c$, it must satisfy $f(c) \ge f(x)$ for all $x$ in the domain. For a piecewise function, this implies that the value at the point of interest must be greater than or equal to the limit from the left and the limit from the right at that point.
• Continuity at a Point: For a piecewise function to be continuous at a point $c$ where the definition changes, the value of the function at $c$ must be equal to the limit from the left and the limit from the right at $c$. That is, $f(c) = \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$.
• Logarithm Properties: For $\log_b a$ to be defined, $a > 0$ and $b > 0, b \ne 1$. The inequality $\log_b a \le k$ implies $a \le b^k$ if $b > 1$, and $a \ge b^k$ if $0 < b < 1$.

Step-by-Step Solution

Step 1: Define the function and the condition for maximum value. The given function is: f(x) = \left\{ {\matrix{ {{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr { - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr } } \right. For $f(x)$ to have a maximum value at $x = 1$, we must have $f(1) \ge f(x)$ for all $x$ in the domain. This implies two conditions:

1. $f(1) \ge \lim_{x \to 1^-} f(x)$
2. $f(1) \ge \lim_{x \to 1^+} f(x)$

Step 2: Calculate $f(1)$. Since $x \le 1$ for the first piece of the function, we use the first expression to find $f(1)$: $f(1) = (1)^3 - (1)^2 + 10(1) - 7 = 1 - 1 + 10 - 7 = 3$

Step 3: Calculate the limit from the left, $\lim_{x \to 1^-} f(x)$. As $x$ approaches 1 from the left ($x \le 1$), we use the first expression of $f(x)$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^3 - x^2 + 10x - 7)$ Since this is a polynomial, the limit can be found by direct substitution: $\lim_{x \to 1^-} f(x) = (1)^3 - (1)^2 + 10(1) - 7 = 1 - 1 + 10 - 7 = 3$ The condition $f(1) \ge \lim_{x \to 1^-} f(x)$ becomes $3 \ge 3$, which is always true. This indicates that the function is continuous from the left at $x=1$.

Step 4: Calculate the limit from the right, $\lim_{x \to 1^+} f(x)$. As $x$ approaches 1 from the right ($x > 1$), we use the second expression of $f(x)$: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (-2x + \log_2(b^2 - 4))$ Since $x$ approaches 1, we substitute $x=1$ into the expression: $\lim_{x \to 1^+} f(x) = -2(1) + \log_2(b^2 - 4) = -2 + \log_2(b^2 - 4)$

Step 5: Apply the condition $f(1) \ge \lim_{x \to 1^+} f(x)$. We have $f(1) = 3$ and $\lim_{x \to 1^+} f(x) = -2 + \log_2(b^2 - 4)$. So, the condition for a maximum at $x=1$ becomes: $3 \ge -2 + \log_2(b^2 - 4)$ Adding 2 to both sides: $5 \ge \log_2(b^2 - 4)$ $\log_2(b^2 - 4) \le 5$

Step 6: Solve the logarithmic inequality. For the logarithm $\log_2(b^2 - 4)$ to be defined, the argument must be positive: $b^2 - 4 > 0 \quad (*)$ Now, consider the inequality $\log_2(b^2 - 4) \le 5$. Since the base of the logarithm is 2 (which is greater than 1), we can exponentiate both sides with base 2 without changing the direction of the inequality: $b^2 - 4 \le 2^5$ $b^2 - 4 \le 32$ Adding 4 to both sides: $b^2 \le 36$ Taking the square root of both sides, we get: $|b| \le 6$ This means $-6 \le b \le 6$.

Step 7: Solve the domain condition for the logarithm. From step 6, we have the condition $b^2 - 4 > 0$. $b^2 > 4$ Taking the square root of both sides: $|b| > 2$ This means $b < -2$ or $b > 2$. In interval notation, this is $b \in (-\infty, -2) \cup (2, \infty)$.

Step 8: Find the intersection of the solutions from Step 6 and Step 7. We need to satisfy both $b^2 \le 36$ (which is $-6 \le b \le 6$) and $b^2 - 4 > 0$ (which is $b \in (-\infty, -2) \cup (2, \infty)$). Let's visualize this on a number line:

• The interval for $b^2 \le 36$ is $[-6, 6]$.
• The interval for $b^2 - 4 > 0$ is $(-\infty, -2) \cup (2, \infty)$.

The intersection of these two sets is: $[-6, 6] \cap ((-\infty, -2) \cup (2, \infty))$ This gives us two separate intervals: $[-6, -2)$ and $(2, 6]$.

Therefore, the set of all values of $b$ for which $f(x)$ has a maximum value at $x=1$ is $[-6, -2) \cup (2, 6]$.

Common Mistakes & Tips

• Forgetting the domain of the logarithm: Always ensure that the argument of a logarithm is strictly positive. In this case, $b^2 - 4 > 0$ is a crucial condition that must be satisfied.
• Incorrectly handling inequalities with logarithms: When solving $\log_b a \le k$, remember that if $0 < b < 1$, the inequality direction reverses when you exponentiate. Here, the base is 2, so the direction remains the same.
• Confusing strict and non-strict inequalities: The condition for the logarithm's argument is strict ($>0$), while the result of the logarithmic inequality may include endpoints ($\le$). Ensure the final answer reflects these distinctions.

Summary

To find the values of $b$ for which $f(x)$ has a maximum at $x=1$, we first evaluated $f(1)$ and the limits from the left and right at $x=1$. The condition for a maximum at $x=1$ is $f(1) \ge \lim_{x \to 1^+} f(x)$, as the function is continuous from the left. This led to a logarithmic inequality, $\log_2(b^2 - 4) \le 5$. We also had to consider the domain of the logarithm, $b^2 - 4 > 0$. Solving these inequalities simultaneously, we found that $b^2 \le 36$ and $b^2 > 4$. The intersection of these conditions yields $b \in [-6, -2) \cup (2, 6]$.

The final answer is \boxed{[-6, -2) \cup (2,6]}.