Function Inflection Points

Approach: Differentiate twice, find critical points, and analyze sign changes.

Steps:

1. $f'(x) = 3x^2 - 12x + 9$
2. $f''(x) = 6x - 12 = 6(x-2)$
3. Setting $f''(x) = 0$ gives $x = 2$
4. For $x < 2$, $f''(x) < 0$, so the function is concave
5. For $x > 2$, $f''(x) > 0$, so the function is convex
6. $f(2) = 8 - 24 + 18 + 1 = 3$

Answer:

The inflection point is $(2, 3)$.

Approach: Differentiate twice, find critical points, and analyze sign changes.

Steps:

1. $f'(x) = \frac{3x^2(x^2 + 1) - x^3 \cdot 2x}{(x^2 + 1)^2} = \frac{x^2(x^2 + 3)}{(x^2 + 1)^2}$
2. $f''(x) = \frac{(2x(x^2 + 3) + x^2 \cdot 2x)(x^2 + 1)^2 - x^2(x^2 + 3) \cdot 2(x^2 + 1) \cdot 2x}{(x^2 + 1)^4}$
3. Simplifying: $f''(x) = \frac{2x(3 - x^2)}{(x^2 + 1)^3}$
4. Setting $f''(x) = 0$ gives $x = 0, \pm \sqrt{3}$
5. Analyzing the sign changes:
   • $x = 0$: sign changes from negative to positive, so it is an inflection point
   • $x = \pm \sqrt{3}$: sign changes from positive to negative, so they are inflection points

Answer:

The inflection points are $(0, 0)$, $(\sqrt{3}, \frac{3\sqrt{3}}{4})$, $(-\sqrt{3}, -\frac{3\sqrt{3}}{4})$.

Approach: Differentiate twice, find critical points, and analyze sign changes.

Steps:

1. $f'(x) = \cos x$
2. $f''(x) = -\sin x$
3. Setting $f''(x) = 0$ gives $\sin x = 0$, i.e. $x = k\pi$ ($k$ an integer)
4. Analyzing sign changes:
   • As $x$ moves from $(2k-1)\pi$ to $2k\pi$, $f''(x)$ changes from positive to negative
   • As $x$ moves from $2k\pi$ to $(2k+1)\pi$, $f''(x)$ changes from negative to positive
5. Therefore every $x = k\pi$ is an inflection point

Answer:

The inflection points are $(k\pi, 0)$, where $k$ is an integer.

Approach: First use the limit to determine $f(1)$ and $f'(1)$, then check whether the hypotheses are enough to fix $f''(1)$ and its sign on both sides.

Steps:

1. From $\lim\limits_{x\to1} \frac{f(x)}{\ln x}=1$, we have $f(x)\sim \ln x$ as $x\to1$
2. As $x\to1$, $\ln x\to0$, so $\lim\limits_{x\to1} f(x)=0$, i.e. $f(1)=0$
3. By the definition of the derivative: $$f'(1)=\lim_{x\to1}\frac{f(x)-f(1)}{x-1} =\lim_{x\to1}\frac{f(x)}{\ln x}\cdot\frac{\ln x}{x-1}=1$$
4. But the hypotheses are not enough to determine the second derivative. Take $$f_c(x)=\ln x+c(x-1)^2\quad (x>0),\qquad f_c(1)=0,$$ which satisfies the hypotheses for every constant $c$, with $f_c''(1)=2c-1$.
5. For example, when $c=0$, $f_c''(x)=-1/x^2<0$, so $x=1$ is not an inflection point; when $c=\frac12$, $f_c''(x)=1-1/x^2$ changes sign around $x=1$, so $x=1$ is an inflection point.

Answer:

From the given hypotheses we cannot determine whether $x=1$ is an inflection point.

Approach: Use the parametric differentiation formulas and analyze the sign change of the second derivative.

Steps:

1. For $t\geq0$: $x=3t$, $y=t\sin t$, so $y=\frac{x}{3}\sin\frac{x}{3}$
2. For $t<0$: $x=t$, $y=-t\sin t$, so $y=-x\sin x$
3. Hence $f(x)=\begin{cases}\frac{x}{3}\sin\frac{x}{3}, & x\geq0\\-x\sin x, & x<0\end{cases}$
4. First derivative: $f'(x)=\begin{cases}\frac{1}{3}\sin\frac{x}{3}+\frac{x}{9}\cos\frac{x}{3}, & x\geq0\\-\sin x-x\cos x, & x<0\end{cases}$
5. Second derivative: $f''(x)=\begin{cases}\frac{2}{9}\cos\frac{x}{3}-\frac{x}{27}\sin\frac{x}{3}, & x\geq0\\-2\cos x+x\sin x, & x<0\end{cases}$
6. At $x=0$:
   • Left derivative: $f''_-(0)=-2\cos 0+0\cdot\sin 0=-2$
   • Right derivative: $f''_+(0)=\frac{2}{9}\cos 0-0\cdot\sin 0=\frac{2}{9}$
7. Since the left and right second derivatives differ, $f''(0)$ does not exist
8. Sign change:
   • For $x<0$, $f''(x)<0$ (concave)
   • For $x>0$, $f''(x)>0$ (convex)

Answer:

$x=0$ is an inflection point of $f(x)$.

Approach: Differentiate the integral with respect to its upper limit, twice, and then track the sign of $f''(x)$ on the two sides of $x = 0$.

Steps:

1. By the fundamental theorem of calculus, $f'(x) = e^{\cos x}$.
2. Differentiating once more, $f''(x) = -e^{\cos x}\sin x$.
3. At $x = 0$: $f'(0) = e^{\cos 0} = e$ and $f''(0) = -e^{\cos 0}\sin 0 = 0$, so the necessary condition is satisfied.
4. The factor $e^{\cos x}$ is positive for every $x$, so the sign of $f''(x)$ is the opposite of the sign of $\sin x$:
   • For $-\delta < x < 0$ we have $\sin x < 0$, hence $f''(x) > 0$ and the graph is convex there.
   • For $0 < x < \delta$ we have $\sin x > 0$, hence $f''(x) < 0$ and the graph is concave there.
5. The second derivative therefore changes sign as $x$ passes through $0$, which is precisely the first sufficient condition.

Answer:

$x = 0$ is an inflection point of $f(x)$; the corresponding point on the graph is $(0, f(0)) = (0, 0)$.

Approach: Differentiate the integral twice and test the necessary condition: if a function is twice differentiable at a point, that point can be an inflection point only when the second derivative there vanishes.

Steps:

1. By the fundamental theorem of calculus, $f'(x) = e^{x^2}\sin x$.
2. Differentiating once more with the product rule, $f''(x) = 2x e^{x^2}\sin x + e^{x^2}\cos x$.
3. At $x = 0$: $f'(0) = e^{0^2}\sin 0 = 0$ and $f''(0) = 2 \cdot 0 \cdot e^{0^2}\sin 0 + e^{0^2}\cos 0 = 1$.
4. Since $f''(0) = 1 \neq 0$, the necessary condition fails and no sign analysis of $f''$ is needed.

Answer:

$x = 0$ is not an inflection point of $f(x)$, because $f''(0) = 1 \neq 0$.

Approach: The three point values only verify the necessary condition. An inflection point requires a sign change of $f''$ on a whole neighborhood, so look for two functions that share all the given data but behave differently around $x = 1$.

Steps:

1. Because $f''(1) = 0$, the necessary condition for an inflection point holds, but the data say nothing about the sign of $f''$ away from $1$.
2. Take $f_1(x) = (x-1) + (x-1)^4$. Then $f_1(1) = 0$, $f_1'(1) = 1$, $f_1''(1) = 0$, and $f_1''(x) = 12(x-1)^2 \geq 0$ on both sides of $x = 1$: the sign does not change, so $x = 1$ is not an inflection point.
3. Take $f_2(x) = (x-1) + \dfrac{1}{3}(x-1)^3$. It satisfies the same three conditions, but $f_2''(x) = 2(x-1)$ is negative just left of $1$ and positive just right of $1$, so $x = 1$ is an inflection point.
4. Both functions meet every hypothesis, so the hypotheses alone cannot decide the question.

Answer:

The data are insufficient. They give only $f''(1) = 0$, the necessary condition, and the two examples above show that $x = 1$ may or may not be an inflection point.

Approach: Again the point values only meet the necessary condition. Test two simple functions that share the given data and compare the sign of their second derivatives near $0$.

Steps:

1. From $g''(0) = 0$ we get the necessary condition, but the sign of $g''(x)$ near $0$ is not determined.
2. For $g_1(x) = x^3$: $g_1(0) = 0$, $g_1'(0) = 0$, $g_1''(0) = 0$, and $g_1''(x) = 6x$ changes sign at $0$, so $x = 0$ is an inflection point.
3. For $g_2(x) = x^4$: $g_2(0) = 0$, $g_2'(0) = 0$, $g_2''(0) = 0$, and $g_2''(x) = 12x^2 \geq 0$ on both sides of $0$, so $x = 0$ is not an inflection point.
4. Since both functions satisfy all the hypotheses, the question has no unique answer.

Answer:

It cannot be decided from the given information: $g(x) = x^3$ and $g(x) = x^4$ show that $x = 0$ may or may not be an inflection point.

Approach: Use the definition of $f'''(0)$ as the derivative of $f''$ at $0$. It says that $f''(x)$ behaves like $f'''(0)\cdot x$ near $0$, so $f''$ must take opposite signs on the two sides of $0$.

Steps:

1. Since $f$ is three times differentiable, $f''$ is differentiable at $0$; using $f''(0) = 0$ we get $$f'''(0)=\lim_{x\to0}\frac{f''(x)-f''(0)}{x-0}=\lim_{x\to0}\frac{f''(x)}{x}\neq0.$$
2. Put $L = f'''(0)$. The limit above means that for all sufficiently small $x \neq 0$ the quotient $f''(x)/x$ lies within $\dfrac{|L|}{2}$ of $L$, so $$\frac{f''(x)}{x}\cdot L > 0.$$ In other words, $f''(x)$ has the same sign as $x\cdot L$.
3. Hence there is $\delta > 0$ such that:
   • if $L > 0$ then $f''(x) < 0$ on $(-\delta, 0)$ and $f''(x) > 0$ on $(0, \delta)$;
   • if $L < 0$ then $f''(x) > 0$ on $(-\delta, 0)$ and $f''(x) < 0$ on $(0, \delta)$. Either way $f''$ changes sign as $x$ passes through $0$.
4. A differentiable function is continuous, so $f$ is continuous on a neighborhood of $0$. The sign change of $f''$ at $0$ is exactly the first sufficient condition, and therefore $(0, f(0))$ is an inflection point.

Answer:

The proof is complete: $x = 0$ is an inflection point of $f(x)$, because $f'''(0) \neq 0$ forces $f''$ to change sign at $0$.