Let f(x) = x³ - x² + x + 1 and g(x) = max {f(t); 0 ≤ t ≤ x}, 0 ≤ x ≤ 1 = 3-x, 1

Q: Wait, why does the $\max$ function just become $f(x)$? I thought it was supposed to be more complicated.

That is a brilliant observation! The $\max$ function is only complicated if the function $f(t)$ oscillates (goes up and down). If $f(t)$ is strictly increasing, then for any interval $[0, x]$, the largest value will always be at the rightmost endpoint, which is $f(x)$. Since we proved $f'(x) > 0$ for all $x$, the function is always climbing. Thus, the maximum is always the value at the current point $x$.

Q: Why do we only check $x = 1$ for continuity and differentiability? What about the rest of the interval $(0, 2)$?

Excellent question. Within the sub-intervals $(0, 1)$ and $(1, 2)$, the function $g(x)$ is defined by simple polynomials ($x^3 - x^2 + x + 1$ and $3 - x$, respectively). Polynomials are inherently continuous and differentiable everywhere. The only 'danger zone' is the junction point $x = 1$, where the definition of the function switches. That is where we must verify if the two pieces 'glue' together smoothly.

Q: If the function is continuous at $x = 1$, why isn't it automatically differentiable?

Continuity is a necessary condition for differentiability, but it is not sufficient. Think of it like this: continuity means the graph is connected (no jumps), but differentiability means the graph is 'smooth' (no sharp corners). You can have a continuous function with a sharp corner, like the absolute value function $|x|$ at $x = 0$. In our case, the slopes of the two pieces at $x = 1$ are different, creating a 'kink' or sharp corner, which prevents the derivative from existing at that point.

Question

Let $f(x) = x^3 - x^2 + x + 1$ and $g(x) = \begin{cases} \max \{f(t); 0 \leq t \leq x\}, & 0 \leq x \leq 1 \ 3-x, & 1 < x \leq 2 \end{cases}$. Discuss the continuity and differentiability of the function $g(x)$ in the interval $(0, 2)$.

Sigma Percentile · Accepted Answer

### The Mystery of the Piecewise Function

Welcome, my dear student. Today, we are going to dissect a problem that often strikes fear into the hearts of JEE aspirants: the piecewise function involving a $\max$ operator.

It looks intimidating, doesn't it? $g(x) = \max \{f(t); 0 \leq t \leq x\}$. But I want you to take a deep breath.

In mathematics, as in life, the most complex-looking problems often hide a simple, elegant truth. Let us peel back the layers together.

### Phase 1: The Monotonicity Investigation

Our first task is to demystify the $\max$ function. The expression $\max \{f(t); 0 \leq t \leq x\}$ asks us to find the largest value of $f(t)$ as $t$ ranges from $0$ to $x$.

If $f(t)$ were to wiggle up and down, this would be a nightmare. But what if $f(t)$ is always climbing?

If $f(t)$ is strictly increasing, then for any interval $[0, x]$, the largest value must occur at the rightmost point, $t = x$. Therefore, the expression simplifies to $f(x)$.

To confirm this, we must check the monotonicity of $f(x) = x^3 - x^2 + x + 1$. We do this by finding its derivative:

$$f'(x) = \frac{d}{dx}(x^3 - x^2 + x + 1) = 3x^2 - 2x + 1$$

Now, we have a quadratic expression $3x^2 - 2x + 1$. How do we know if this is always positive? We look at the discriminant $D = b^2 - 4ac$:

$$D = (-2)^2 - 4(3)(1) = 4 - 12 = -8$$

Since $D < 0$ and the leading coefficient ($3$) is positive, the parabola $f'(x)$ never touches the x-axis and stays entirely above it. This means $f'(x) > 0$ for all $x$.

Our function $f(x)$ is **strictly increasing**! The mystery of the $\max$ function is solved: for $0 \leq x \leq 1$, $g(x) = f(x) = x^3 - x^2 + x + 1$.

### Phase 2: The Junction Point

Now that we have simplified $g(x)$ to a standard piecewise function, we must examine the junction point $x = 1$. This is where the two definitions meet: $x^3 - x^2 + x + 1$ for $x \leq 1$ and $3 - x$ for $x > 1$.

To check for continuity, we compare the left-hand limit (LHL), the right-hand limit (RHL), and the function value $g(1)$:

$$LHL = \lim_{x 	o 1^-} (x^3 - x^2 + x + 1) = 1^3 - 1^2 + 1 + 1 = 2$$

$$RHL = \lim_{x 	o 1^+} (3 - x) = 3 - 1 = 2$$

$$g(1) = 1^3 - 1^2 + 1 + 1 = 2$$

Since $LHL = RHL = g(1)$, the function is **perfectly continuous** at $x = 1$. The graph is connected.

### Phase 3: The Sharp Corner

Finally, we address differentiability. Continuity is not enough; we need smoothness. We must check if the slopes match at $x = 1$.

We calculate the left-hand derivative (LHD) and the right-hand derivative (RHD):

$$LHD = g'(1^-) = \frac{d}{dx}(x^3 - x^2 + x + 1) \Big|_{x=1} = (3x^2 - 2x + 1) \Big|_{x=1} = 3(1)^2 - 2(1) + 1 = 2$$

$$RHD = g'(1^+) = \frac{d}{dx}(3 - x) \Big|_{x=1} = -1$$

Look at that! The LHD is $2$, but the RHD is $-1$. The slopes do not match.

This means that at $x = 1$, the graph takes a sharp turn. It is continuous, but it is **not differentiable**.

### Conclusion

We have navigated the complexity of the $\max$ function, verified the continuity, and uncovered the sharp corner at $x = 1$.

The final verdict is that $g(x)$ is continuous on $(0, 2)$, but differentiable only on $(0, 2) \setminus \{1\}$. You have successfully mastered this problem.

Remember, in JEE, it is not just about the calculation; it is about the conceptual journey. Keep practicing, and keep falling in love with the logic behind the math!

Animated Solution for Mathematics - Differentiation: Let $f (x) = x^{3} - x^{2} + x + 1$ and $g (x) = {max {f (t); 0 \leq t \leq x}, 3 - x, 0 \leq x \leq 1 1 < x \leq 2$ . Discuss the continuity and differentiability of the function $g (x)$ in the interval $(0, 2)$ .

Visualized Solution

Understanding the Function $g (x)$

Analyzing Monotonicity of $f (x)$

Differentiating $f (x)$

Checking the Discriminant of $f^{'} (x)$

Conclusion on Monotonicity

Simplifying $g (x)$ on $[0, 1]$

Checking Continuity at $x = 1$

Evaluating Limits at $x = 1$

Checking Differentiability at $x = 1$

Evaluating Derivatives at $x = 1$

Final Conclusion

The Sigma Insight: Relationship Between Continuity and Differentiability

The Mystery of the Piecewise Function

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

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Visualized Solution

Understanding the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Analyzing Monotonicity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Checking the Discriminant of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Conclusion on Monotonicity

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

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

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

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