If f(x) = {3x²+12x-1, -1≤x≤2; 37-x, 2

Q: Wait, why is the derivative at $x=2$ undefined even though the function is continuous?

This is a classic JEE trap! Continuity only guarantees that the graph is connected, not that it is smooth. A function can be continuous but have a 'sharp corner' or 'cusp' where the slope changes abruptly. At $x=2$, the slope from the left is $24$, but the slope from the right is $-1$. Since the slopes don't match, the derivative cannot exist at that point.

Q: Can I just check the derivative at $x=2$ by plugging it into the quadratic part?

No, that would only give you the Left-Hand Derivative (LHD). Because the function is defined differently on either side of $x=2$, you must calculate the derivative for both the quadratic piece and the linear piece separately. The derivative at a point exists only if both one-sided derivatives are equal.

Q: How do I know if it's increasing on the whole interval $[-1, 2]$?

To check for increasing behavior, we look at the sign of the derivative $f'(x)$. For the quadratic part, $f'(x) = 6x + 12$. On the interval $[-1, 2]$, the smallest value of $x$ is $-1$, which gives $f'(-1) = 6$. Since the derivative is always positive ($f'(x) \ge 6$) throughout the interval, the function must be strictly increasing.

Question

If $f(x) = \begin{cases} 3x^2+12x-1, & -1\le x\le 2 \ 37-x, & 2<x\le 3 \end{cases}$ then:

Options:

(A) $f(x)$ is increasing on $[-1, 2]$

(B) $f(x)$ is continuous on $[-1, 3]$

(C) $f'(2)$ does not exist

(D) $f(x)$ has the maximum value at $x = 2$

Sigma Percentile · Accepted Answer

### The Art of the Piecewise Function: A Journey Through Continuity and Change

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a function that, at first glance, might seem like a simple puzzle, but it holds the keys to understanding the fundamental nature of calculus: continuity and differentiability.

We are looking at the function defined as:

$$f(x) = \begin{cases} 3x^2+12x-1, & -1 \le x \le 2 \ 37-x, & 2 < x \le 3 \end{cases}$$

### Phase 1: The Parabolic Ascent

Imagine you are standing at $x = -1$. You are about to walk along the path defined by $f(x) = 3x^2 + 12x - 1$.

As you move towards $x = 2$, we calculate the derivative:

$$f'(x) = 6x + 12$$

For any $x$ in our interval $[-1, 2]$, the smallest value $6x$ can take is $6(-1) = -6$. Adding $12$ to this gives us a minimum slope of $6$.

Since the slope is always positive, you are constantly climbing. This is the beauty of a strictly increasing function; there is no looking back, no turning around. You start at $f(-1) = -10$ and reach a peak of $f(2) = 35$.

### Phase 2: The Linear Descent

Now, imagine you reach the summit at $x = 2$. Suddenly, the rules change.

You are no longer on a parabola; you are now walking along a straight line defined by $f(x) = 37 - x$. This is a line with a constant slope of $-1$.

As you move from $x = 2$ to $x = 3$, you are descending. The function value drops from $35$ to $34$.

### Phase 3: The Junction at $x=2$

This is the moment of truth. We check for continuity by comparing the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL) at $x = 2$.

The LHL, approaching from the parabola, is:

$$LHL = 3(2)^2 + 12(2) - 1 = 35$$

The RHL, approaching from the line, is:

$$RHL = 37 - 2 = 35$$

They match, meaning the graph is continuous and there is no gap. However, continuity is not the same as smoothness.

We check the derivatives: the Left-Hand Derivative (LHD) is $f'(2^-) = 6(2) + 12 = 24$, while the Right-Hand Derivative (RHD) is $f'(2^+) = -1$.

Because $24 
eq -1$, the slope changes instantly from a steep climb to a gentle descent. This creates a sharp corner, or a cusp, in the graph.

### Phase 4: The Verdict

We have analyzed the function's behavior, its continuity, and its differentiability. We found that it is strictly increasing on $[-1, 2]$, continuous on $[-1, 3]$, and that it possesses a sharp corner at $x = 2$, meaning the derivative does not exist there.

Finally, we identified the **absolute maximum value of $35$ at $x = 2$**.

This problem is a perfect reminder that in the world of JEE, you must look beyond the surface. Always check the boundaries, always verify the derivatives, and never assume smoothness without proof.

Animated Solution for Mathematics - Differentiation: If $f (x) = {3 x^{2} + 12 x - 1, 37 - x, - 1 \leq x \leq 2 2 < x \leq 3$ then:

Select Answer:

Visualized Solution

Visualizing the Piecewise Function

Plotting $f (x) = 3 x^{2} + 12 x - 1$

Monotonicity on $[- 1, 2]$

Plotting the Linear Part

Checking Continuity at $x = 2$

Left-Hand Derivative (LHD) at $x = 2$

Right-Hand Derivative (RHD) at $x = 2$

Sharp Corner at $x = 2$

Finding the Maximum Value

Verifying the Correct Options

The Sigma Insight: Monotonicity

The Art of the Piecewise Function

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

Let $f : R \to R$ be defined by $f (x) = \frac{x ^{2} - 3 x - 6}{x ^{2} + 2 x + 4}$ . Then which of the following statements is (are) TRUE ?

If $f (x) = \int_{x^{2}}^{x^{2} + 1} e^{- t^{2}} d t$ , then $f (x)$ increases in

If $f : R \to R$ is a differentiable function such that $f^{'} (x) > 2 f (x)$ for all $x \in R$ , and $f (0) = 1$ , then

The function, $f (x) = (3 x - 7) x^{2/3}$ , $x \in R$ , is increasing for all $x$ lying in :

For the function $f (x) = sin x + 3 x - \frac{2}{π} (x^{2} + x)$ , where $x \in [0, \frac{π}{2}]$ , consider the following two statements : (I) $f$ is increasing in $(0, \frac{π}{2})$ . (II) $f^{'}$ is decreasing in $(0, \frac{π}{2})$ . Between the above two statements,

The function $f (x) = \frac{4 x ^{3} - 3 x ^{2}}{6} - 2 sin x + (2 x - 1) cos x$ :

If $f (x) = x^{3} + b x^{2} + c x + d$ and $0 < b^{2} < c$ , then in $(- \infty, \infty)$

The function defined by $f (x) = (x + 2) e^{- x}$ is

Let $f$ be a real valued function, defined on $R - {- 1, 1}$ and given by $f (x) = 3 lo g_{e} \frac{x - 1}{x + 1} - \frac{2}{x - 1}$ . Then in which of the following intervals, function $f (x)$ is increasing?

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)={3x2+12x−1,37−x,​−1≤x≤22<x≤3​ then:

Select Answer:

Visualized Solution

Visualizing the Piecewise Function

Plotting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=3x2+12x−1

Monotonicity on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [−1,2]

Plotting the Linear Part

Checking Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

Left-Hand Derivative (LHD) at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

Right-Hand Derivative (RHD) at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

Sharp Corner at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2