Question

Let $f(x) = \begin{cases} x+1, & 0 \leq x \leq 1 \\ 2x^2 - 6x + 6, & 1 < x \leq 2 \end{cases}$ and $g(t) = \int_{t-1}^{t} f(x) dx$ for $t \in [1,2]$

Which of the following hold(s) good?

• A. $f(x)$ is continuous and differentiable in [0, 2]
• B. $g'(t)$ vanishes for t = 3/2 and 2
• C. $g(t)$ is maximum at t = 3/2
• D. $g(t)$ is minimum at t = 1

Answer 1

Answer: (B), (C), (D)

B, C, D

Answer 2

1. Analysis of $f(x)$ for Continuity and Differentiability (Option A)

The function $f(x)$ is defined as: $f(x) = \begin{cases} x+1, & 0 \leq x \leq 1 \\ 2x^2 - 6x + 6, & 1 < x \leq 2 \end{cases}$

• Continuity at $x=1$:

  • Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+1) = 1+1 = 2$.
  • Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x^2 - 6x + 6) = 2(1)^2 - 6(1) + 6 = 2 - 6 + 6 = 2$.
  • Function value at $x=1$: $f(1) = 1+1 = 2$. Since the left-hand limit, right-hand limit, and function value are equal, $f(x)$ is continuous at $x=1$. As both pieces are polynomials, $f(x)$ is continuous on $[0, 2]$.

• Differentiability at $x=1$:

  • Left-hand derivative: $f'(x) = \frac{d}{dx}(x+1) = 1$ for $0 \leq x < 1$. So, $f'(1^-) = 1$.
  • Right-hand derivative: $f'(x) = \frac{d}{dx}(2x^2 - 6x + 6) = 4x - 6$ for $1 < x \leq 2$. So, $f'(1^+) = 4(1) - 6 = 4 - 6 = -2$. Since $f'(1^-) \neq f'(1^+)$, $f(x)$ is not differentiable at $x=1$. Therefore, Option A is false.

2. Analysis of $g(t)$ and its derivative $g'(t)$

$g(t) = \int_{t-1}^{t} f(x) dx$ for $t \in [1,2]$. Using the Leibniz integral rule, $g'(t) = \frac{d}{dt} \int_{u(t)}^{v(t)} f(x) dx = f(v(t))v'(t) - f(u(t))u'(t)$. Here, $v(t) = t$ and $u(t) = t-1$. So, $v'(t) = 1$ and $u'(t) = 1$. $g'(t) = f(t) \cdot 1 - f(t-1) \cdot 1 = f(t) - f(t-1)$.

We need to consider the definition of $f(x)$ based on the value of $x$. For $t \in [1, 2]$:

• The argument $t$ is in $[1, 2]$.
• The argument $t-1$ is in $[0, 1]$.

So, for $t \in [1, 2]$:

• $f(t-1) = (t-1)+1 = t$ (since $0 \leq t-1 \leq 1$).
• For $f(t)$:
  • If $t=1$, $f(1) = 1+1 = 2$.
  • If $1 < t \leq 2$, $f(t) = 2t^2 - 6t + 6$.

Let's evaluate $g'(t)$:

• For $t=1$: $g'(1) = f(1) - f(0) = (1+1) - (0+1) = 2 - 1 = 1$.
• For $1 < t \leq 2$: $g'(t) = (2t^2 - 6t + 6) - t = 2t^2 - 7t + 6$.

3. Analysis of Option B

Option B states that $g'(t)$ vanishes for $t = 3/2$ and $2$. We examine $g'(t)$ for $1 < t \leq 2$: Set $g'(t) = 0 \implies 2t^2 - 7t + 6 = 0$. Factoring the quadratic equation: $(2t - 3)(t - 2) = 0$. The roots are $t = 3/2$ and $t = 2$. Both $3/2$ and $2$ are in the interval $(1, 2]$. Also, $g'(1) = 1 \neq 0$. Thus, $g'(t)$ vanishes for $t=3/2$ and $t=2$. Option B is true.

4. Analysis of Options C and D

To analyze where $g(t)$ is maximum or minimum, we study the sign of $g'(t) = 2t^2 - 7t + 6$ for $t \in [1, 2]$. The roots of $g'(t)$ are $3/2$ and $2$. The parabola $y = 2t^2 - 7t + 6$ opens upwards.

• For $t \in [1, 3/2)$: $g'(t) > 0$ (e.g., $g'(1.1) = 2(1.1)^2 - 7(1.1) + 6 = 2.42 - 7.7 + 6 = 0.72 > 0$). This means $g(t)$ is increasing.
• For $t \in (3/2, 2)$: $g'(t) < 0$ (e.g., $g'(1.6) = 2(1.6)^2 - 7(1.6) + 6 = 5.12 - 11.2 + 6 = -0.08 < 0$). This means $g(t)$ is decreasing.
• At $t=3/2$, $g'(3/2) = 0$.

This indicates that $g(t)$ has a local maximum at $t=3/2$. To determine if it's a global maximum, we evaluate $g(t)$ at critical points and endpoints. Let's calculate $g(t)$ for $t \in [1, 2]$: $g(t) = \int_{t-1}^{1} (x+1) dx + \int_{1}^{t} (2x^2 - 6x + 6) dx$ $g(t) = \left[\frac{x^2}{2} + x\right]_{t-1}^{1} + \left[\frac{2x^3}{3} - 3x^2 + 6x\right]_{1}^{t}$ $g(t) = \left(\frac{1}{2}+1\right) - \left(\frac{(t-1)^2}{2} + (t-1)\right) + \left(\frac{2t^3}{3} - 3t^2 + 6t\right) - \left(\frac{2}{3} - 3 + 6\right)$ $g(t) = \frac{3}{2} - \frac{t^2-2t+1+2t-2}{2} + \frac{2t^3}{3} - 3t^2 + 6t - \frac{11}{3}$ $g(t) = \frac{3}{2} - \frac{t^2-1}{2} + \frac{2t^3}{3} - 3t^2 + 6t - \frac{11}{3}$ $g(t) = \frac{4-t^2}{2} + \frac{2t^3}{3} - 3t^2 + 6t - \frac{11}{3}$ $g(t) = 2 - \frac{t^2}{2} + \frac{2t^3}{3} - 3t^2 + 6t - \frac{11}{3}$ $g(t) = \frac{2t^3}{3} - \frac{7t^2}{2} + 6t - \frac{5}{3}$

Now evaluate $g(t)$ at $t=1, 3/2, 2$:

• $g(1) = \frac{2(1)^3}{3} - \frac{7(1)^2}{2} + 6(1) - \frac{5}{3} = \frac{2}{3} - \frac{7}{2} + 6 - \frac{5}{3} = -1 - \frac{7}{2} + 6 = 5 - \frac{7}{2} = \frac{3}{2} = 1.5$.
• $g(3/2) = \frac{2(3/2)^3}{3} - \frac{7(3/2)^2}{2} + 6(3/2) - \frac{5}{3} = \frac{2(27/8)}{3} - \frac{7(9/4)}{2} + 9 - \frac{5}{3} = \frac{9}{4} - \frac{63}{8} + 9 - \frac{5}{3} = \frac{54-189+216-40}{24} = \frac{41}{24} \approx 1.708$.
• $g(2) = \frac{2(2)^3}{3} - \frac{7(2)^2}{2} + 6(2) - \frac{5}{3} = \frac{16}{3} - 14 + 12 - \frac{5}{3} = \frac{11}{3} - 2 = \frac{5}{3} \approx 1.667$.

Comparing the values: $g(1) = 1.5$, $g(3/2) \approx 1.708$, $g(2) \approx 1.667$.

• The maximum value of $g(t)$ in $[1, 2]$ is $41/24$, which occurs at $t=3/2$. So, Option C is true.
• The minimum value of $g(t)$ in $[1, 2]$ is $3/2$, which occurs at $t=1$. So, Option D is true.

Conclusion: Options B, C, and D are correct.