12. Suppose $$ f, g: \mathbb{R} \rightarrow \mathbb{R} $$, that $$ f(x)=(x-2)^{2} g(x) $$ for all $$ x \in \mathbb{R} $$, and that $$ g $$ is NOT differentiable at $$ x=2 $$ but is continuous at $$ x=2 $$. Compute $$ f^{\prime}(2) $$ with full justification or explain why $$ f^{\prime}(2) $$ does not exist.

Question

UpStudy AI · Accepted Answer

We are given
$$
f(x) = (x-2)^2 g(x)
$$
and that $$g$$ is continuous at $$x=2$$ (so $$g(2)$$ exists) but not differentiable at $$x=2$$. To compute $$f'(2)$$, we use the definition of the derivative:
$$
f'(2) = \lim_{x \to 2} \frac{f(x) - f(2)}{x-2}.
$$

**Step 1. Compute $$f(2)$$.**

$$
f(2) = (2-2)^2 g(2) = 0^2 g(2) = 0.
$$

**Step 2. Write the difference quotient.**

$$
\frac{f(x) - f(2)}{x-2} = \frac{(x-2)^2 g(x) - 0}{x-2} = (x-2) \, g(x).
$$

**Step 3. Take the limit as $$x \to 2$$.**

$$
f'(2) = \lim_{x \to 2} (x-2) \, g(x).
$$
Since $$g$$ is continuous at $$x=2$$, we have $$\lim_{x \to 2} g(x) = g(2)$$. Therefore,
$$
f'(2) = \left(\lim_{x \to 2} (x-2)\right) \cdot g(2) = 0 \cdot g(2) = 0.
$$

**Conclusion:**

The derivative $$f'(2)$$ exists and is equal to 0.

Even though $$g$$ is not differentiable at $$x=2$$, its continuity is sufficient to ensure the limit exists when multiplied by the factor $$(x-2)$$ which tends to 0.
$$ f'(2) = 0 $$.

Suppose , that for all , and that is NOT differentiable at but is continuous at . Compute with full justification or explain why does not exist.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Calculus Questions