AP DSC School Assistant Math Model Test 3

Concept:The function is given piecewise. We check continuity at x=0 by verifying that the limit as x→0 equals f(0). Then we check differentiability by verifying that the left-hand derivative (LHD) equals the right-hand derivative (RHD). Finally, we examine the derivative function f′(x) to see if it is continuous at x=0.Explanation:1. Continuity at x=0:We need $\lim_{x \to 0} f(x) = f(0)$. For $x \neq 0$, $f(x)=x^{2}\sin(\frac{1}{x})$. Since $|\sin(\frac{1}{x})| \le 1$, we have $|x^{2}\sin(\frac{1}{x})| \le x^{2} \to 0$ as $x \to 0$. Hence $\lim_{x \to 0} f(x)=0$. Also $f(0)=0$. So the limit equals the function value, thus $f$ is continuous at $x=0$.2. Differentiability at x=0:Compute LHD at $x=0$:$\displaystyle f'_{L}(0)=\lim_{h \to 0} \frac{f(0-h)-f(0)}{-h} = \lim_{h \to 0} \frac{(-h)^{2}\sin(\frac{1}{-h})-0}{-h} = \lim_{h \to 0} \frac{h^{2}\sin(-\frac{1}{h})}{-h} = \lim_{h \to 0} \frac{-h^{2}\sin(\frac{1}{h})}{-h} = \lim_{h \to 0} h \sin(\frac{1}{h}) = 0$ (since $|h\sin(1/h)| \le |h| \to 0$).RHD at $x=0$:$\displaystyle f'_{R}(0)=\lim_{h \to 0} \frac{f(0+h)-f(0)}{h} = \lim_{h \to 0} \frac{h^{2}\sin(\frac{1}{h})}{h} = \lim_{h \to 0} h \sin(\frac{1}{h}) = 0$.Since LHD = RHD = 0, $f$ is differentiable at $x=0$ and $f'(0)=0$.3. Continuity of f′(x) at x=0:For $x \neq 0$, differentiate $f(x)$ using product and chain rules:$f'(x) = 2x \sin\left(\frac{1}{x}\right) + x^{2} \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^{2}}\right) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)$.At $x=0$, $f'(0)=0$.Now check the limit of $f'(x)$ as $x \to 0$: $\lim_{x \to 0} f'(x) = \lim_{x \to 0} \left[2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)\right]$. The term $2x\sin(1/x) \to 0$ (bounded times zero), but $\cos(1/x)$ oscillates between -1 and 1 without approaching any limit. Therefore $\lim_{x \to 0} f'(x)$ does not exist. Since the limit does not equal $f'(0)=0$, $f'$ is not continuous at $x=0$.Answer:Option D: f is continuous but f′ is not continuous.