To analyze the function \( f: [0,1] \to \mathbb{R} \) defined as:
\[
f(x) =
\begin{cases}
x^3(1-x) \sin\left(\frac{1}{x^2}\right) & \text{if } 0 < x \leq 1 \\
0 & \text{if } x = 0
\end{cases}
\]
we will check the continuity and differentiability of \( f \) on the interval \([0, 1]\).
### Step 1: Check Continuity at \( x = 0 \)
To determine if \( f \) is continuous at \( x = 0 \), we need to check if:
\[
\lim_{x \to 0^+} f(x) = f(0)
\]
Calculating the limit:
\[
\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^3(1-x) \sin\left(\frac{1}{x^2}\right)
\]
Since \( \sin\left(\frac{1}{x^2}\right) \) oscillates between -1 and 1, we can bound \( f(x) \):
\[
-x^3(1-x) \leq f(x) \leq x^3(1-x)
\]
As \( x \to 0 \):
\[
-x^3(1-x) \to 0 \quad \text{and} \quad x^3(1-x) \to 0
\]
By the Squeeze Theorem:
\[
\lim_{x \to 0^+} f(x) = 0
\]
Thus, \( f(0) = 0 \) and \( \lim_{x \to 0^+} f(x) = f(0) \). Therefore, \( f \) is continuous at \( x = 0 \).
### Step 2: Check Continuity on \( (0, 1] \)
For \( 0 < x \leq 1 \), \( f(x) \) is composed of continuous functions (polynomials and sine function). Hence, \( f \) is continuous on \( (0, 1] \).
Combining both parts, \( f \) is continuous on the entire interval \([0, 1]\).
### Step 3: Check Differentiability at \( x = 0 \)
To check differentiability at \( x = 0 \), we compute:
\[
f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h}
\]
Substituting \( f(h) \):
\[
f'(0) = \lim_{h \to 0} \frac{h^3(1-h) \sin\left(\frac{1}{h^2}\right)}{h} = \lim_{h \to 0} h^2(1-h) \sin\left(\frac{1}{h^2}\right)
\]
Again, using the bounds on the sine function:
\[
-h^2(1-h) \leq h^2(1-h) \sin\left(\frac{1}{h^2}\right) \leq h^2(1-h)
\]
As \( h \to 0 \):
\[
-h^2(1-h) \to 0 \quad \text{and} \quad h^2(1-h) \to 0
\]
By the Squeeze Theorem:
\[
\lim_{h \to 0} h^2(1-h) \sin\left(\frac{1}{h^2}\right) = 0
\]
Thus, \( f'(0) = 0 \).
### Step 4: Check Differentiability on \( (0, 1] \)
For \( 0 < x \leq 1 \), \( f(x) \) is differentiable since it is composed of differentiable functions. Therefore, \( f \) is differentiable on \( (0, 1] \).
### Conclusion
- \( f \) is continuous on \([0, 1]\).
- \( f \) is differentiable on \((0, 1]\) and \( f'(0) = 0 \).
### Final Answers
- (a) False: \( f \) is differentiable at \( x = 0 \).
- (b) True: \( f \) is continuous on \([0, 1]\).
- (c) True: \( f \) is bounded on \([0, 1]\) as it is continuous on a closed interval.
- (d) False: \( f' \) is not bounded as \( x \to 0 \).
