Continuity and Differentiability of Integral Defined Function

To analyze the continuity and differentiability of the function $g(x)$, we consider the given definition:

$$g(x) = \begin{cases} 0 & \text{if } x < a \\ \int_a^x f(t) \, dt & \text{if } a \le x \le b \\ \int_a^b f(t) \, dt & \text{if } x > b \end{cases}$$

where $f: [a, b] \rightarrow [1, \infty)$ is a continuous function.

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1. Continuity Analysis

• At $x = a$:

  • Left-Hand Limit (LHL): $\lim_{x \to a^-} g(x) = \lim_{x \to a^-} 0 = 0$
  • Right-Hand Limit (RHL): $\lim_{x \to a^+} g(x) = \lim_{x \to a^+} \int_a^x f(t) \, dt = \int_a^a f(t) \, dt = 0$
  • Value of the function at $x = a$: $g(a) = \int_a^a f(t) \, dt = 0$ Since $\text{LHL} = \text{RHL} = g(a) = 0$, $g(x)$ is continuous at $x = a$.

• At $x = b$:

  • Left-Hand Limit (LHL): $\lim_{x \to b^-} g(x) = \lim_{x \to b^-} \int_a^x f(t) \, dt = \int_a^b f(t) \, dt$
  • Right-Hand Limit (RHL): $\lim_{x \to b^+} g(x) = \lim_{x \to b^+} \int_a^b f(t) \, dt = \int_a^b f(t) \, dt$
  • Value of the function at $x = b$: $g(b) = \int_a^b f(t) \, dt$ Since $\text{LHL} = \text{RHL} = g(b)$, $g(x)$ is continuous at $x = b$.

• Everywhere else:

  • For $x < a$, $g(x) = 0$ is a constant function, hence continuous.
  • For $a < x < b$, $g(x) = \int_a^x f(t) \, dt$ is continuous by the Fundamental Theorem of Calculus, as $f(t)$ is continuous on $[a, b]$.
  • For $x > b$, $g(x) = \int_a^b f(t) \, dt$ is a constant, hence continuous.

Thus, $g(x)$ is continuous everywhere on $\mathbb{R}$.

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2. Differentiability Analysis

• At $x = a$:

  • Left-Hand Derivative (LHD): $g'(a^-) = \lim_{h \to 0^-} \frac{g(a+h) - g(a)}{h} = \lim_{h \to 0^-} \frac{0 - 0}{h} = 0$
  • Right-Hand Derivative (RHD): By the Fundamental Theorem of Calculus, $g'(a^+) = \lim_{x \to a^+} \frac{d}{dx} \left( \int_a^x f(t) \, dt \right) = f(a)$ Since $f: [a, b] \rightarrow [1, \infty)$, we have $f(a) \ge 1 > 0$. Therefore, $g'(a^-) \neq g'(a^+)$, which implies that $g(x)$ is not differentiable at $x = a$.

• At $x = b$:

  • Left-Hand Derivative (LHD): By the Fundamental Theorem of Calculus, $g'(b^-) = \lim_{x \to b^-} \frac{d}{dx} \left( \int_a^x f(t) \, dt \right) = f(b)$
  • Right-Hand Derivative (RHD): $g'(b^+) = \lim_{h \to 0^+} \frac{g(b+h) - g(b)}{h} = \lim_{h \to 0^+} \frac{\int_a^b f(t) \, dt - \int_a^b f(t) \, dt}{h} = 0$ Since $f(b) \ge 1 > 0$, we have $g'(b^-) \neq g'(b^+)$, which implies that $g(x)$ is not differentiable at $x = b$.

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Conclusion

• $g(x)$ is continuous everywhere on $\mathbb{R}$.
• $g(x)$ is not differentiable at $x = a$ and $x = b$.

Hence, options (A) and (C) are correct.