Let f : R → R be defined as f(x) = { -55x, if x

Q: Why do we need to check continuity at the boundaries $x = -5$ and $x = 4$?

Great question! While monotonicity is determined by the derivative, a jump discontinuity at a boundary could technically affect the definition of 'increasing' across the entire domain. Checking continuity ensures our piecewise pieces are 'glued' together, allowing us to treat the union of intervals as a valid solution set. It is a safety check to ensure the function behaves predictably at the transition points.

Q: In the second piece, the derivative is positive for $x > 5$ and $x < -4$. Why is my answer only $(-5, -4)$?

Ah, the classic trap! You found the roots correctly, but you forgot the domain constraint. The second piece is only defined for $-5 \leq x \leq 4$. You must intersect your result $(-\infty, -4) \cup (5, \infty)$ with the interval $[-5, 4]$. The only overlapping region is $(-5, -4)$. Always respect the boundaries of the piece you are currently analyzing!

Q: Why is the first piece $f(x) = -55x$ not increasing?

Look at the slope! The derivative is $-55$. Since $-55 < 0$, the function is strictly decreasing. A negative slope means the function is sliding downhill, not climbing. Always check the sign of the derivative, even for simple linear pieces. Never assume a function is increasing just because it looks simple.

Question

Let $f : R ightarrow R$ be defined as
$f(x) = \begin{cases} -55x, & 	ext{if } x < -5 \ 2x^3 - 3x^2 - 120x, & 	ext{if } -5 \leq x \leq 4 \ 2x^3 - 3x^2 - 36x - 336, & 	ext{if } x > 4 \end{cases}$
Let $A = \{x \in R : f 	ext{ is increasing } \}$. Then $A$ is equal to :

Options:

(A) $(-5, -4) \cup (4,\infty)$

(B) $(-5, \infty)$

(C) $(-\infty, -5) \cup (4,\infty)$

(D) $(-\infty, -5) \cup (-4,\infty)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Welcome, fellow traveler of the mathematical realm. Today, we are not just solving a problem; we are dissecting a 'Frankenstein' function. Piecewise functions are like stories told in chapters, where each chapter has its own rules and trajectory. Our goal is to find where this function is **increasing**—where it is climbing uphill—by analyzing each chapter individually using calculus. ### Chapter 1: The Linear Descent We begin our journey in the domain $x < -5$. Here, the function is defined as $f(x) = -55x$. Differentiating this linear path, we obtain: $$f'(x) = -55$$ A derivative of $-55$ indicates that for every step to the right, the function drops by $55$ units. Since $-55 < 0$, the function is strictly decreasing here. There is no climbing to be found in this chapter. ### Chapter 2: The Cubic Rollercoaster Next, we enter the middle chapter: $-5 \leq x \leq 4$. The function is defined as $f(x) = 2x^3 - 3x^2 - 120x$. To understand the slope, we differentiate: $$f'(x) = 6x^2 - 6x - 120$$ To find where the function is increasing, we set $f'(x) > 0$. Factoring out the common term $6$, we get: $$6(x^2 - x - 20) > 0$$ $$6(x - 5)(x + 4) > 0$$ Using the wavy curve method, the quadratic is positive in the intervals $(-\infty, -4) \cup (5, \infty)$. However, we must intersect this with our specific domain $[-5, 4]$. The only portion of our domain that falls into the 'positive' region is the interval $(-5, -4)$. This is our first victory; in this small window, the function is climbing. ### Chapter 3: The Final Ascent Finally, we reach the last chapter: $x > 4$. The function is defined as $f(x) = 2x^3 - 3x^2 - 36x - 336$. Differentiating this expression, we obtain: $$f'(x) = 6x^2 - 6x - 36$$ Factoring this, we get: $$6(x^2 - x - 6) = 6(x - 3)(x + 2)$$ The critical points are $x = -2$ and $x = 3$. The derivative is positive for $x < -2$ and $x > 3$. Since our domain for this piece is $x > 4$, and the entire interval $(4, \infty)$ is a subset of $(3, \infty)$, the derivative is positive for the entire duration of this chapter. The function is climbing steadily towards infinity. ### The Synthesis We have traversed all three chapters. In the first, we found only descent. In the second, we found a brief climb in $(-5, -4)$. In the third, we found a continuous climb in $(4, \infty)$. By combining these findings, we arrive at our final set: **$$A = (-5, -4) \cup (4, \infty)$$** This problem teaches us a vital lesson: never assume. By breaking the function down, respecting the domain boundaries, and rigorously applying the derivative test, we have uncovered the truth behind the function's behavior.

Animated Solution for Mathematics - Differentiation: Let $f : R \to R$ be defined as\n $f (x) = ⎩ ⎨ ⎧ - 55 x, 2 x^{3} - 3 x^{2} - 120 x, 2 x^{3} - 3 x^{2} - 36 x - 336, if x < - 5 if - 5 \leq x \leq 4 if x > 4$ \nLet $A = {x \in R : f is increasing}$ . Then $A$ is equal to :

Select Answer:

Visualized Solution

Understanding Monotonicity

Differentiating Piece 1: $x < - 5$

Behavior of Piece 1

Differentiating Piece 2: $- 5 \leq x \leq 4$

Factorizing the Derivative

Sign Analysis for Piece 2

Decreasing Region of Piece 2

Differentiating Piece 3: $x > 4$

Factorizing the Third Derivative

Sign Analysis for Piece 3

Checking Continuity

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

Chapter 1

Chapter 2

Chapter 3

The Synthesis

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