Q.Find the maximum and minimum values, if any, of the following functions given by

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

Concept: Quadratic Extrema (applied to absolute value — a V‑shaped function).

The absolute value $|x+2|$ is minimum ($0$) when $x = -2$.

Thus $f(x)_{\min} = 0 - 1 = -1$.

As $|x+2|$ can grow without bound, $f(x)$ has no maximum.

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

Concept: Inverted V‑shape — maximum at the vertex.

$|x+1|$ is minimum ($0$) at $x = -1$, so $g(x)_{\max} = -0 + 3 = 3$.

Since $|x+1|$ can increase arbitrarily, $g(x)$ has no minimum.

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

Concept: Sine function range $[-1, 1]$.

Minimum: $-1 + 5 = 4$; Maximum: $1 + 5 = 6$.

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

Concept: Absolute value of a shifted sine.

$\sin 4x \in [-1, 1] \Rightarrow \sin 4x + 3 \in [2, 4]$, both positive.

So $f(x) = \sin 4x + 3$ itself, range $[2, 4]$.

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

Concept: Open interval — no endpoints attained.

As $x \to -1^+$, $h(x) \to 0$ (but never equals $0$).

As $x \to 1^-$, $h(x) \to 2$ (but never equals $2$).

No absolute maximum or minimum exist.

Concept and Intuition

The key to finding extrema of these functions is to identify the range of the inner expression first.

• For absolute value functions $|u|$, the smallest value is $0$ (when $u=0$), and there is no upper bound unless $u$ itself is bounded.
• For $\sin(2x)$, the range is $[-1, 1]$.
• For a linear function on an open interval, the endpoints are not attained, so no absolute maximum or minimum exists — only supremum and infimum.

Once we know the range of the inner part, we apply the outer operation (adding a constant, taking absolute value, etc.) to get the range of the whole function. The extremum is then read directly from that range.

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

1. Inner expression: $|x+2|$ is always $\ge 0$, and it attains $0$ when $x = -2$. It can become arbitrarily large as $x \to \pm\infty$.

2. Outer operation: Subtract $1$. So $f(x) = |x+2| - 1 \ge -1$.

3. Minimum: Achieved when $|x+2| = 0$, i.e., at $x = -2$. Then $f(-2) = 0 - 1 = -1$.

4. Maximum: Since $|x+2|$ has no upper bound, $f(x)$ can be made arbitrarily large. Hence no maximum.

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

1. Inner expression: $|x+1| \ge 0$, with equality at $x = -1$.

2. Outer operation: The negative sign flips the V upside down: $-|x+1| \le 0$. Then add $3$: $g(x) \le 3$.

3. Maximum: Achieved when $|x+1| = 0$, i.e., at $x = -1$. Then $g(-1) = 0 + 3 = 3$.

4. Minimum: As $|x+1| \to \infty$, $-|x+1| \to -\infty$, so $g(x) \to -\infty$. No minimum.

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

1. Inner expression: $\sin(2x)$ has range $[-1, 1]$, and it attains both endpoints infinitely often (e.g., at $x = \pi/4$ for $1$, $x = 3\pi/4$ for $-1$).

2. Outer operation: Add $5$. So $h(x)$ ranges from $-1+5 = 4$ to $1+5 = 6$.

3. Minimum: $4$, attained when $\sin(2x) = -1$.

4. Maximum: $6$, attained when $\sin(2x) = 1$.

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

1. Inner expression: $\sin 4x$ ranges from $-1$ to $1$. So $\sin 4x + 3$ ranges from $2$ to $4$.

2. Outer operation: Absolute value. Since the entire range $[2,4]$ is already non-negative, the absolute value does nothing — it just keeps the same range.

3. Minimum: $2$, attained when $\sin 4x = -1$.

4. Maximum: $4$, attained when $\sin 4x = 1$.

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

1. Domain: Open interval $(-1, 1)$. The function is strictly increasing.

2. Range: As $x$ approaches $-1$ from the right, $h(x) \to 0$; as $x$ approaches $1$ from the left, $h(x) \to 2$. But neither endpoint is included in the domain.

3. Extrema:

   • No minimum: $0$ is a lower bound but never attained.
   • No maximum: $2$ is an upper bound but never attained.
   • The function has no absolute maximum or minimum on this open interval.

Method: Finding Extrema by Bounding the Inner Expression's Range

Many "find the max/min" questions are not calculus problems at all — they are range problems. Whenever the function is built from a piece with a known, fixed range (absolute value, sine, cosine) combined with a simple outer operation (add/subtract a constant, take absolute value, negate), the fastest and most reliable method is to bound the inner piece first and propagate that bound outward.

Steps

Step 1: Identify the "core" bounded expression and its natural range.

• $|u| \in [0, \infty)$, with the minimum $0$ attained when $u=0$.
• $\sin(\text{anything}) \in [-1, 1]$ and $\cos(\text{anything}) \in [-1, 1]$, both bounds attained infinitely often.

Step 2: Apply the outer operation to the range, term by term.

If $f(x) = |u(x)| + k$, then since $|u(x)| \ge 0$, $f(x) \ge k$, with equality exactly when $u(x)=0$ — so $f$ has a minimum $k$ and no maximum (because $|u(x)|$ is unbounded above whenever $u(x)$ itself is unbounded). If instead $f(x) = -|u(x)| + k$, the inequality flips: $f(x) \le k$, giving a maximum $k$ and no minimum. For a sine/cosine core, $a\sin(\cdots) + b$ ranges over $[b-|a|, b+|a|]$ directly.

Step 3: Check whether the whole transformed range is already one-signed before applying a further absolute value.

If an outer $|\cdot|$ is wrapped around a range that is already entirely $\ge 0$ (or entirely $\le 0$), the absolute value changes nothing — the range stays exactly as computed in Step 2.

Step 4 (Applying to this problem): For a function restricted to an open interval, check attainability separately.

Even a perfectly monotonic function like $x+1$ has bounds it approaches on an open interval $(a,b)$ — but since the endpoints themselves are excluded from the domain, those bounding values are never actually reached, so there is no absolute maximum or minimum, only a supremum/infimum.

Common Mistakes

Mistake 1: Assuming a "V-shaped" absolute-value graph has both a maximum and a minimum.

Why it's wrong: $|x+2|-1$ opens upward without bound as $x\to\pm\infty$, so only the vertex (the minimum) is a real extremum — there is no maximum to report. Correct approach: check which direction the V opens (a plain $|u|$ opens up, a $-|u|$ opens down) and only claim the extremum that direction actually produces.

Mistake 2: Reporting the open-interval endpoint values as the maximum and minimum.

Why it's wrong: for $h(x)=x+1$ on $(-1,1)$, the values $0$ and $2$ are only limits the function approaches as $x\to -1$ and $x\to 1$ — since those endpoints are excluded from the domain, $h(x)$ never actually equals $0$ or $2$. Correct approach: state explicitly that no absolute maximum or minimum exists whenever the domain is an open interval and the function is monotonic on it.

Mistake 3: Applying the outer absolute value in part (iv) as if it could lower the range.

Why it's wrong: since $\sin 4x+3$ already lies entirely in $[2,4]$ (all positive), wrapping it in $|\cdot|$ does nothing — a student who reflexively assumes $|\cdot|$ must change the minimum to $0$ gets a wrong answer. Correct approach: always check the sign of the pre-absolute-value range before deciding whether $|\cdot|$ has any effect.