A function is matched below against an interval where it is supposed to be increasing. Which of the

Q: Wait, why do we use $f'(x) \ge 0$ and not just $f'(x) > 0$?

Great question! In calculus, a function is considered 'increasing' on an interval if $f(x_2) \ge f(x_1)$ for all $x_2 > x_1$. This allows for the derivative to be zero at isolated points (like the vertex of a parabola). So, $f'(x) \ge 0$ is the correct, rigorous condition for a function to be non-decreasing, which is what we mean by 'increasing' in this context.

Q: Can I just pick a point in the interval and check the sign of the derivative?

Yes, you can! Testing a point is a fantastic way to verify your work. For example, in option (c), if you pick $x = 0$ (which is in the interval $(-\infty, 1/3]$), the derivative $f'(0) = 6(0) - 2 = -2$. Since the derivative is negative, the function is decreasing at that point, which immediately tells you the interval is not an increasing one.

Q: Why is the interval $(-\infty, 1/3]$ decreasing for $3x^2 - 2x + 1$?

Because the derivative $f'(x) = 6x - 2$ is negative for all $x < 1/3$. When the derivative is negative, the slope of the tangent is downward, meaning the function values are getting smaller as $x$ increases. That is the definition of a decreasing function!

Question

A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched? \ 	extbf{Interval} \quad 	extbf{Function} \ (a) $(-\infty, \infty) \quad x^3 - 3x^2 + 3x + 3$ \ (b) $[2, \infty) \quad 2x^3 - 3x^2 - 12x + 6$ \ (c) $(-\infty, 1/3] \quad 3x^2 - 2x + 1$ \ (d) $(-\infty, -4) \quad x^3 + 6x^2 + 6$

Options:

(A) $(-\infty, \infty), x^3 - 3x^2 + 3x + 3$

(B) $[2, \infty), 2x^3 - 3x^2 - 12x + 6$

(C) $(-\infty, 1/3], 3x^2 - 2x + 1$

(D) $(-\infty, -4), x^3 + 6x^2 + 6$

Sigma Percentile · Accepted Answer

### The Detective Work of Calculus: Unmasking Monotonicity

Imagine you are standing on a mountain path. To know if you are climbing, you don't need to see the whole mountain; you just need to look at the slope of the ground right beneath your feet.

If the ground tilts upward, you are increasing your altitude. If it tilts downward, you are decreasing. In the world of functions, the derivative $f'(x)$ is exactly that slope.

### The Tool: The Derivative as a Compass

Before we dive into the options, let's sharpen our primary tool. A differentiable function $f(x)$ is increasing on an interval if its derivative, $f'(x)$, is greater than or equal to zero for all $x$ in that interval.

This is our mathematical compass. If $f'(x) \ge 0$, the graph is heading uphill. If $f'(x) \le 0$, it is heading downhill.

### Analyzing the Options

Let's start with Option (a): $f(x) = x^3 - 3x^2 + 3x + 3$. The derivative is:

$$f'(x) = 3x^2 - 6x + 3$$

If we factor out the $3$, we get $3(x^2 - 2x + 1)$, which is $3(x-1)^2$. Since any real number squared is non-negative, $3(x-1)^2 \ge 0$ for all $x \in \mathbb{R}$. This function is always climbing, so Option (a) is a correct match.

Now, look at Option (b): $f(x) = 2x^3 - 3x^2 - 12x + 6$. Its derivative is:

$$f'(x) = 6x^2 - 6x - 12$$

Factoring out the $6$, we get $6(x^2 - x - 2)$, which splits into $6(x-2)(x+1)$. For the interval $[2, \infty)$, both $(x-2)$ and $(x+1)$ are positive. A positive times a positive is positive, so $f'(x) \ge 0$. This function is indeed increasing, making Option (b) a correct match.

### The Trap in Option (c)

Now, let's turn our attention to the culprit: Option (c). The function is $f(x) = 3x^2 - 2x + 1$. The derivative is:

$$f'(x) = 6x - 2$$

To find where it is increasing, we set $6x - 2 \ge 0$, which simplifies to $x \ge \frac{1}{3}$. This means the function is increasing on the interval $[\frac{1}{3}, \infty)$.

But look at what the option claims! It says the function is increasing on $(-\infty, \frac{1}{3}]$. If we test a value in that interval, say $x = 0$, we get $f'(0) = -2$.

The derivative is negative! The function is actually decreasing on this interval. We have caught our culprit red-handed.

### Final Verification

Just to be absolutely certain, let's check Option (d): $f(x) = x^3 + 6x^2 + 6$. The derivative is:

$$f'(x) = 3x^2 + 12x = 3x(x+4)$$

For the interval $(-\infty, -4)$, both $x$ and $(x+4)$ are negative. The product of two negative numbers is positive, so $f'(x) > 0$. This function is indeed increasing, confirming that Option (d) is also a correct match.

By systematically checking the derivative of each function, we have unmasked the mismatch. **Option (c)** was the only one that failed the test, as it paired a decreasing interval with a claim of increase. Keep this detective mindset, and no calculus problem will ever be able to hide its secrets from you!

Select Answer:

Visualized Solution

The Objective

The Monotonicity Condition

Checking Option (a)

Checking Option (b)

Analyzing Option (c)

The Critical Point

True Increasing Interval

True Decreasing Interval

The Trap in Option (c)

The Mismatch

Verifying Option (d)

Final Conclusion

The Sigma Insight: Monotonicity

The Detective Work of Calculus

The Tool

Analyzing the Options

The Trap in Option (c)

Final Verification

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List-II

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Select Answer:

Visualized Solution

The Objective

The Monotonicity Condition

Checking Option (a)

Checking Option (b)

Analyzing Option (c)

The Critical Point

True Increasing Interval

True Decreasing Interval

The Trap in Option (c)

The Mismatch

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The Tool

Analyzing the Options

The Trap in Option (c)

Final Verification

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