Question8. Let $f:[a, b] \rightarrow \mathbb{R}$ be continuous on $[a, b]$ and differenchable in $(a, b)$. Show that if $\lim _{x \rightarrow a} f^{\prime}(x)=A$, then $f^{\prime}(a)$ exists and equals $A$. [Hint: Use the definition of $f^{\prime}(a)$ and the Mean Value Theorem.]
9. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x):=2 x^{4}+x^{4} \sin (1 / x)$ for $x \neq 0$ and $f(0):=0$. Show that $f$ has an absolute minimum at $x=0$, but that its derivative has both positive and negative values in every neighborhood of 0 .

STEP 1

1. $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$.
2. $\lim_{x \rightarrow a} f^{\prime}(x) = A$.
3. We need to show $f^{\prime}(a)$ exists and equals $A$.
_HIGH_LEVEL_APPROACH_ for Problem 8:
1. Use the definition of the derivative at a point.
2. Apply the Mean Value Theorem.
3. Show that the derivative at $a$ exists and equals $A$.

STEP 2

STEP 3

Use the definition of the derivative at a point:
The derivative $f^{\prime}(a)$ is defined as:
$f^{\prime}(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

STEP 4

Apply the Mean Value Theorem:
For any $x$ in $(a, b)$, there exists a $c$ in $(a, x)$ such that:
$f^{\prime}(c) = \frac{f(x) - f(a)}{x - a}$

STEP 5

Show that the derivative at $a$ exists and equals $A$:
Given $\lim_{x \rightarrow a} f^{\prime}(x) = A$, for any $\epsilon > 0$, there exists $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f^{\prime}(x) - A| < \epsilon$.
Using the Mean Value Theorem, as $x \to a$, $c \to a$ and thus $\lim_{c \to a} f^{\prime}(c) = A$.
Therefore, $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = A$, so $f^{\prime}(a) = A$.
_HIGH_LEVEL_APPROACH_ for Problem 9:
1. Determine the value of $f(0)$.
2. Show that $f(x) \geq f(0)$ for all $x$.
3. Analyze the derivative $f^{\prime}(x)$ in neighborhoods of 0.
4. Show that the derivative takes both positive and negative values near 0.
STEP_1: Determine the value of $f(0)$:
Given $f(0) = 0$.
STEP_2: Show that $f(x) \geq f(0)$ for all $x$:
For $x \neq 0$, $f(x) = 2x^4 + x^4 \sin(1/x)$.
Since $-1 \leq \sin(1/x) \leq 1$, it follows that:
$x^4(-1) \leq x^4 \sin(1/x) \leq x^4(1)$
Thus:
$x^4 \geq x^4 \sin(1/x) \geq -x^4$
Therefore:
$2x^4 + x^4 \sin(1/x) \geq 0$
Hence, $f(x) \geq f(0)$.
STEP_3: Analyze the derivative $f^{\prime}(x)$ in neighborhoods of 0:
Calculate the derivative:
For $x \neq 0$:
$f^{\prime}(x) = 8x^3 + 4x^3 \sin(1/x) - x^2 \cos(1/x)$

STEP 6

Show that the derivative takes both positive and negative values near 0:
Consider the term $-x^2 \cos(1/x)$. As $x \to 0$, $\cos(1/x)$ oscillates between $-1$ and $1$, causing $-x^2 \cos(1/x)$ to oscillate between $-x^2$ and $x^2$.
Thus, $f^{\prime}(x)$ can take both positive and negative values in every neighborhood of 0.
The function $f$ has an absolute minimum at $x = 0$, and its derivative has both positive and negative values in every neighborhood of 0.