Let f(x) = { x³ - x² + 10x - 7, x ≤ 1; -2x + log₂(b² - 4), x 1 }. Then the set of all values of b,

Q: Wait, why do we use the inequality $\le$ instead of just $<$ for the limit condition?

Great question! A local maximum at $x = 1$ means that $f(1)$ must be greater than or equal to all values in a small neighborhood around $1$. If the right-hand limit were strictly less than $f(1)$, that would be fine, but it could also be equal to $f(1)$ if the function is continuous or if the jump is downwards. The condition for a maximum is simply that no point in the neighborhood is higher than $f(1)$.

Q: Why do we need to check the domain of the logarithm? Isn't the inequality enough?

The inequality only tells us where the function *could* be a maximum, but the function must exist in the first place! The expression $\log_2(b^2 - 4)$ is only defined if $b^2 - 4 > 0$. If we ignore this, we might include values of $b$ that make the function undefined, which would mean there is no maximum at all.

Q: Why is the final interval $[-6, -2) \cup (2, 6]$? Why are the brackets different?

This is a classic trap! The closed brackets at $-6$ and $6$ come from the inequality $b^2 \le 36$, which includes the endpoints. However, the open brackets at $-2$ and $2$ come from the domain constraint $b^2 > 4$. Since the function is undefined at $b = 2$ and $b = -2$, we must exclude those points, hence the open intervals.

Question

Let $f(x) = \begin{cases} x^3 - x^2 + 10x - 7, & x \le 1 \ -2x + \log_2(b^2 - 4), & x > 1 \end{cases}$. Then the set of all values of $b$, for which $f(x)$ has maximum value at $x = 1$, is :

Options:

(A) $(-6, -2)$

(B) $(2, 6)$

(C) $[-6, -2) \cup (2, 6]$

(D) $[-\sqrt{6}, -2) \cup (2, \sqrt{6}]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To determine if the landmark at $x = 1$ acts as a local maximum for the piecewise function, we must ensure that the function reaches its peak at this point and does not exceed this value in the immediate neighborhood.

The function is defined as $f(x) = x^3 - x^2 + 10x - 7$ for $x \le 1$. By substituting $x = 1$ into this cubic expression, we find the elevation of our anchor point:

$$f(1) = 1^3 - 1^2 + 10(1) - 7 = 3$$

Thus, our anchor point is located at the coordinate $(1, 3)$. This value serves as the threshold that the rest of the path must not surpass.

### The Climb (The Left Branch)

We analyze the behavior of the path as we approach the peak from the left by calculating the derivative of the cubic branch:

$$f'(x) = 3x^2 - 2x + 10$$

To determine the nature of this slope, we examine the discriminant $D$ of the quadratic derivative:

$$D = (-2)^2 - 4(3)(10) = 4 - 120 = -116$$

Since the discriminant is negative and the leading coefficient is positive, the derivative $f'(x)$ is always positive. This confirms that for $x < 1$, the function is strictly increasing, meaning we are consistently climbing toward our peak at $(1, 3)$.

### The Descent (The Right Branch)

For $x > 1$, the function is defined by the linear expression $f(x) = -2x + \log_2(b^2 - 4)$. The derivative of this branch is constant at $-2$.

Because the slope is negative, the function is strictly decreasing for all $x > 1$. This confirms that the path descends immediately after the landmark, which is a necessary condition for a local maximum.

### The Bridge (The Limit Condition)

For $x = 1$ to be a maximum, the function must not jump to a value higher than $3$ as we transition into the right branch. Mathematically, the right-hand limit must satisfy:

$$\lim_{x 	o 1^+} f(x) \le f(1)$$

Substituting our expression, we obtain:

$$-2(1) + \log_2(b^2 - 4) \le 3$$

Adding $2$ to both sides isolates the logarithmic term:

$$\log_2(b^2 - 4) \le 5$$

Converting this to exponential form yields:

$$b^2 - 4 \le 2^5 \Rightarrow b^2 - 4 \le 32 \Rightarrow b^2 \le 36$$

This inequality results in the range $b \in [-6, 6]$.

### The Hidden Trap

We must also ensure the function is well-defined. The term $\log_2(b^2 - 4)$ requires the argument to be strictly positive:

$$b^2 - 4 > 0 \Rightarrow b^2 > 4$$

This implies $|b| > 2$, which corresponds to the interval $b \in (-\infty, -2) \cup (2, \infty)$.

### Final Synthesis

To find the valid set of values for $b$, we intersect our two conditions: $b \in [-6, 6]$ and $b \in (-\infty, -2) \cup (2, \infty)$.

The intersection of these sets provides the final solution:

**$$b \in [-6, -2) \cup (2, 6]$$**

Animated Solution for Mathematics - Differentiation: Let $f (x) = {x^{3} - x^{2} + 10 x - 7, - 2 x + lo g_{2} (b^{2} - 4), x \leq 1 x > 1$ . Then the set of all values of $b$ , for which $f (x)$ has maximum value at $x = 1$ , is :

Select Answer:

Visualized Solution

Evaluate $f (1)$

Analyze $f (x)$ for $x < 1$

Monotonicity for $x < 1$

Analyze $f (x)$ for $x > 1$

Condition for Maximum at $x = 1$

Formulate the Inequality

Solve the Logarithmic Inequality

Remove the Logarithm

Range from Inequality

Logarithm Domain Constraint

Solve Domain Constraint

Final Intersection

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Climb (The Left Branch)

The Descent (The Right Branch)

The Bridge (The Limit Condition)

The Hidden Trap

Final Synthesis

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