Stationary points, convexity and an alternating series¶

Math I: Calculus · Week 6 test, open question · 7 points · Lectures 9–10

Problem¶

Consider $f(x)=x^2(x-2)^2$ for $x\in\mathbb{R}\text{.}$

(a) Find every stationary point of $f$ and classify each one as a local maximum or local minimum. [3 points]
(b) Find the $x$ -coordinates of all inflection points. State the intervals on which $f$ is convex and those on which it is concave. [2 points]
(c) Determine whether the series below converges, and justify your conclusion. [2 points]

$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}$

What you need¶

Stationary point and the second-order condition: $f''(x_0)>0$ gives a local minimum, $f''(x_0)<0$ a local maximum.
Higher-order derivative.
Convexity and concavity and inflection point: $f''$ must change sign, not only vanish.
Alternating series test.

Plan¶

Factor $f'$ completely before solving $f'=0\text{;}$ the factored form shows the three stationary points at once. Then $f''$ does two jobs: its value at each stationary point classifies it, and its zeros and sign give the inflection points and the convexity intervals. Part (c) is independent: check the two conditions of the alternating series test.

Step-by-step solution¶

Try each step yourself before you open it.

Step 1 · Compute and factor the derivatives (counted in (a) and (b))

By the product rule,

\begin{aligned} f'(x)&=2x(x-2)^2+2x^2(x-2)\\ &=2x(x-2)(2x-2)\\ &=4x(x-1)(x-2). \end{aligned}

Expanding $f'(x)=4x^3-12x^2+8x$ and differentiating again,

f''(x)=12x^2-24x+8.

(a) · Classify the stationary points (3 points)

$f'(x)=0$ exactly at $x=0,1,2\text{.}$ Then

f''(0)=8,\quad f''(1)=-4,\quad f''(2)=8.

By the second-order condition, $x=0$ and $x=2$ are local minima and $x=1$ is a local maximum.

Concept: Second-order condition

(b) · Inflection points and convexity (2 points)

$f''(x)=4(3x^2-6x+2)=0$ gives

x=1\pm\frac{1}{\sqrt{3}}.

$f''$ is a parabola opening upwards, so it is positive outside these two roots and negative between them. The sign changes at both roots, so both are inflection points. $f$ is convex on $(-\infty,1-1/\sqrt{3})$ and on $(1+1/\sqrt{3},\infty)\text{,}$ and concave on $(1-1/\sqrt{3},1+1/\sqrt{3})\text{.}$

Concept: Inflection point

(c) · The alternating series (2 points)

Write the series as $\displaystyle \sum(-1)^{n-1}c_n$ with $c_n=1/\sqrt{n}\text{.}$ The sequence $(c_n)$ is positive, decreasing, and tends to $0\text{.}$ By the alternating series test, the series converges.

It does not converge absolutely: $\displaystyle \sum1/\sqrt{n}$ is a $p$ -series with $p=1/2\leq1\text{.}$ The question does not ask for this, but it is the reason the alternating series test is needed.

Concept: Alternating series test

Where marks are lost¶

Calling every zero of $f''$ an inflection point. Check that $f''$ changes sign there.
Classifying by a picture. State the values of $f''$ (or the sign change of $f'$ ) at each stationary point.
Checking only $c_n\to0$ in (c). The test also needs $(c_n)$ to be decreasing.

Try next¶

Symbols: element of, infinity, real numbers, square root, sum