Exercise 6.3 · Q2: Maximum and minimum values of $|x+2|-1$⁠, $-|x+1|+3$⁠, $\sin 2x+5$⁠, $|\sin 4x+3|$⁠, $x+1$ on $(-1,1)$

Basics you need

Part (i): $f(x)=|x+2|-1$

1. Minimum value $-1$⁠, at $x=-2$

   Since $|x+2|\ge0$ for all $x$⁠, subtracting $1$ gives

   $$f(x)=|x+2|-1\ \ge\ -1\quad\text{for all } x\in\R.$$

   Equality needs $|x+2|=0$⁠, i.e. $x=-2$⁠, and $f(-2)=0-1=-1$⁠. So the minimum value is $-1$⁠, at $x=-2$.

   This is Example 15 ($|x|$ has minimum $0$ at $0$⁠) moved $2$ units left and $1$ unit down. Moving a graph does not change its shape, so the corner simply moves to $(-2,-1)$⁠.
   $|t|\ge0$⁠, equality only at $t=0$⁠. Basics: signs

2. No maximum value

   $f(x)=d-1$⁠, where $d=|x+2|$ is the distance from $x$ to $-2$⁠. For any point $c$⁠, move one unit further from $-2$⁠: $x=c+1$ if $c\ge-2$⁠, $x=c-1$ if $c\lt-2$⁠. The distance grows by $1$⁠, so $f(x)=f(c)+1\gt f(c)$⁠. No candidate survives: there is no maximum value.

   The graph is a V whose arms go up for ever. To be completely rigorous we beat an arbitrary $c$⁠, not just "large $x$".

3. Cross-check: a corner, not a zero derivative

   For $x\lt-2$⁠, $f(x)=-(x+2)-1$⁠, so $f'(x)=-1\lt0$⁠. For $x\gt-2$⁠, $f(x)=(x+2)-1$⁠, so $f'(x)=1\gt0$⁠. At $x=-2$⁠, $f$ is not differentiable (the two slopes differ), so $-2$ is a critical point. $f'$ changes from negative to positive there and $f$ is continuous at $-2$⁠, so by the first derivative test (Theorem 3(ii)) $x=-2$ is a point of local minima.

   Theorem 2 says an extreme point inside an open interval has $f'(c)=0$ or is a point where $f$ is not differentiable. Here it is the second kind, exactly as in Example 19 ($3+|x|$⁠).
   Theorem 2, the Note on critical points, Theorem 3, Example 19. Lesson: first and second derivative tests

Part (ii): $g(x)=-|x+1|+3$

1. Maximum value $3$⁠, at $x=-1$

   From $|x+1|\ge0$ we get $-|x+1|\le0$⁠, so

   $$g(x)=3-|x+1|\ \le\ 3\quad\text{for all } x\in\R,$$

   with equality only at $x=-1$⁠, where $g(-1)=-0+3=3$⁠. So the maximum value is $3$⁠, at $x=-1$.

   The minus sign turns the V upside down, so the corner becomes the highest point.
   Multiplying an inequality by $-1$ reverses it. Basics: signs

2. No minimum value

   $g(x)=3-d$ with $d=|x+1|$⁠. For any point $c$⁠, the point one unit further from $-1$ ($c+1$ if $c\ge-1$⁠, $c-1$ if $c\lt-1$⁠) gives the value $g(c)-1\lt g(c)$⁠. So there is no minimum value.

   The arms of the upside-down V fall without end.

3. Cross-check with the first derivative test

   For $x\lt-1$⁠: $g(x)=(x+1)+3$⁠, $g'(x)=1\gt0$⁠. For $x\gt-1$⁠: $g(x)=-(x+1)+3$⁠, $g'(x)=-1\lt0$⁠. $g$ is continuous at $-1$ and $g'$ changes from positive to negative there, so $x=-1$ is a point of local maxima (Theorem 3(i)). ✓

   For $x\lt-1$⁠, $x+1\lt0$ and $|x+1|=-(x+1)$⁠; for $x\gt-1$⁠, $|x+1|=x+1$⁠. That is how each side's formula was written.
   Theorem 3 (First Derivative Test). Lesson: first and second derivative tests

Part (iii): $h(x)=\sin(2x)+5$

1. Bounds from the range of sine

   For every $x$⁠, $-1\le\sin 2x\le1$⁠. Adding $5$ throughout,

   $$4\ \le\ h(x)=\sin 2x+5\ \le\ 6\quad\text{for all } x\in\R.$$
   Whatever the angle $2x$ is, its sine lies in $[-1,1]$⁠. Adding the same number to all three parts of an inequality keeps it true.
   $-1\le\sin t\le1$⁠. Basics: sin and cos on the unit circle

2. Both bounds are reached

   $h(x)=6$ needs $\sin 2x=1$⁠: $2x=\frac{\pi}{2}+2n\pi$⁠, so $x=\frac{\pi}{4}+n\pi$⁠. For example $h\!\left(\frac{\pi}{4}\right)=\sin\frac{\pi}{2}+5=6$⁠.

   $h(x)=4$ needs $\sin 2x=-1$⁠: $2x=\frac{3\pi}{2}+2n\pi$⁠, so $x=\frac{3\pi}{4}+n\pi$⁠. For example $h\!\left(\frac{3\pi}{4}\right)=\sin\frac{3\pi}{2}+5=4$⁠.

   So the maximum value is $6$ and the minimum value is $4$.

   Definition 3 asks for one point where the value is reached; we have infinitely many, one in every stretch of length $\pi$⁠. The value is the same at all of them.
   Definition 3, Section 6.4. Lesson: maxima and minima

3. Cross-check with derivatives

   By the chain rule, $h'(x)=2\cos 2x$ and $h''(x)=-4\sin 2x$⁠. At $x=\frac{\pi}{4}$⁠: $h'=2\cos\frac{\pi}{2}=0$⁠, $h''=-4\lt0$⁠, a point of local maxima. At $x=\frac{3\pi}{4}$⁠: $h'=2\cos\frac{3\pi}{2}=0$⁠, $h''=-4(-1)=4\gt0$⁠, a point of local minima (Theorem 4). ✓

   Inner function $2x$ (derivative $2$⁠), outer function $\sin$ (derivative $\cos$⁠). The derivative test confirms the shape; the range argument already proved the values.
   Theorem 4 (Second Derivative Test). Lesson: tests

Part (iv): $f(x)=|\sin 4x+3|$

1. The absolute value changes nothing here

   From $-1\le\sin 4x\le1$⁠, adding $3$⁠:

   $$2\ \le\ \sin 4x+3\ \le\ 4.$$

   So $\sin 4x+3$ is always positive, and the absolute value of a positive number is the number itself: $f(x)=\sin 4x+3$ for every $x$⁠.

   The modulus only matters when the inside can be negative. Checking the sign first saves us from splitting into cases that never happen.
   $|t|=t$ when $t\ge0$⁠. Basics: signs

2. Both bounds are reached

   $f(x)=4$ when $\sin 4x=1$⁠: $4x=\frac{\pi}{2}+2n\pi$⁠, $x=\frac{\pi}{8}+\frac{n\pi}{2}$⁠. Check: $f\!\left(\frac{\pi}{8}\right)=|\sin\frac{\pi}{2}+3|=4$⁠.

   $f(x)=2$ when $\sin 4x=-1$⁠: $4x=\frac{3\pi}{2}+2n\pi$⁠, $x=\frac{3\pi}{8}+\frac{n\pi}{2}$⁠. Check: $f\!\left(\frac{3\pi}{8}\right)=|-1+3|=2$⁠.

   So the maximum value is $4$ and the minimum value is $2$.

   Dividing $4x=\frac{\pi}{2}+2n\pi$ by $4$ divides the spacing $2n\pi$ by $4$ as well, giving $\frac{n\pi}{2}$⁠.
   $\sin t=\pm1$ at $t=\frac{\pi}{2}+2n\pi$⁠, $\frac{3\pi}{2}+2n\pi$⁠. Basics: sin and cos on the unit circle

Part (v): $h(x)=x+1,\ x\in(-1,1)$

1. The values fill $(0,2)$ but never reach $0$ or $2$

   For $-1\lt x\lt1$⁠, adding $1$ gives $0\lt h(x)\lt2$⁠. The bounds $0$ and $2$ would need $x=-1$ or $x=1$⁠, which are not in the domain.

   This is Example 16 of the book ($f(x)=x$ on $(0,1)$⁠) shifted: a strictly increasing function on an open interval heads towards its end values but the end points are not available.
   Open interval: end points excluded. Basics: interval notation

2. Every candidate point is beaten

   Take any $c\in(-1,1)$⁠.

   • No maximum: the midpoint $x_1=\frac{c+1}{2}$ of $c$ and $1$ lies in $(c,1)$⁠, so it is in the domain and $h(x_1)=x_1+1\gt c+1=h(c)$⁠.
   • No minimum: the midpoint $x_2=\frac{c-1}{2}$ of $-1$ and $c$ lies in $(-1,c)$⁠, so $h(x_2)\lt h(c)$⁠.

   So no point gives the largest value and none gives the smallest: $h$ has neither a maximum nor a minimum value.

   Check the midpoints really are inside: $c\lt\frac{c+1}{2}\lt1$ because $c\lt1$⁠, and $-1\lt\frac{c-1}{2}\lt c$ because $c\gt-1$⁠. Between any point and an excluded end point there is always room for another point.
   Definition 3, Section 6.4; compare Example 16. Lesson: maxima and minima

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