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

(Concept: Quadratic Extrema) A quadratic $ax^2+bx+c$ (or a perfect square form) has a single extremum at its vertex. For $a>0$ the vertex gives a minimum; for $a<0$ it gives a maximum. No global maximum exists if the parabola opens upward, and no global minimum if it opens downward.

(i) $f(x) = (2x-1)^2 + 3$

The square term is $\ge 0$, minimum when $2x-1=0 \Rightarrow x=\frac12$.

Minimum value $= 0+3 = 3$. No maximum (parabola opens up).

(ii) $f(x) = 9x^2 + 12x + 2$

Complete the square: $9\left(x^2 + \frac43 x\right) + 2 = 9\left(x+\frac23\right)^2 - 4 + 2 = 9\left(x+\frac23\right)^2 - 2$.

Minimum at $x=-\frac23$, value $= -2$. No maximum.

(iii) $f(x) = -(x-1)^2 + 10$

Square term $\ge 0$, so $-(x-1)^2 \le 0$. Maximum when $x=1$, value $= 0+10 = 10$. No minimum.

(iv) $g(x) = x^3 + 1$

Cubic function; as $x\to -\infty$, $g(x)\to -\infty$; as $x\to \infty$, $g(x)\to \infty$. No global maximum or minimum.

The Core Idea: Quadratic Extrema

A quadratic function $ax^2 + bx + c$ (with $a \neq 0$) graphs as a parabola. The vertex is the single turning point. If $a > 0$, the parabola opens upward — the vertex gives the minimum value, and the function grows without bound on both sides (no maximum). If $a < 0$, it opens downward — the vertex gives the maximum value, and the function decreases without bound (no minimum).

For a function written in vertex form $f(x) = a(x-h)^2 + k$, the vertex is at $(h, k)$. The extremum value is $k$, occurring at $x = h$. For a function in standard form $ax^2 + bx + c$, the vertex $x$-coordinate is $x = -\frac{b}{2a}$.

Now let's apply this to each part.

(i) $f(x) = (2x-1)^2 + 3$

1. Recognise the form. This is already in vertex form: $f(x) = (2x-1)^2 + 3$. But careful — the squared term is $(2x-1)^2$, not $(x-h)^2$ with a coefficient of 1. Let's rewrite it cleanly.

   Expand: $(2x-1)^2 = 4x^2 - 4x + 1$, so $f(x) = 4x^2 - 4x + 4$. That's $a = 4 > 0$, so the parabola opens upward — only a minimum exists.

2. Find the vertex. For $(2x-1)^2$, the expression inside the square is zero when $2x - 1 = 0$, i.e., $x = \frac12$. At that point, $(2x-1)^2 = 0$, so $f\left(\frac12\right) = 0 + 3 = 3$.

   Since a square is always $\ge 0$, we have $(2x-1)^2 \ge 0$ for all $x$, so $f(x) \ge 3$ for all $x$. The value $3$ is actually attained at $x = \frac12$.

3. Check for a maximum. As $x \to \pm\infty$, $(2x-1)^2 \to \infty$, so $f(x) \to \infty$. There is no upper bound — no maximum.

(ii) $f(x) = 9x^2 + 12x + 2$

1. Identify the shape. Here $a = 9 > 0$, so the parabola opens upward — only a minimum exists.

2. Find the vertex $x$-coordinate. Using $x = -\frac{b}{2a}$:

3. Find the minimum value. Substitute $x = -\frac23$ into $f(x)$:

So the minimum value is $-2$ at $x = -\frac23$.

4. Check for a maximum. As $x \to \pm\infty$, $9x^2$ dominates, so $f(x) \to \infty$. No maximum.

(iii) $f(x) = -(x-1)^2 + 10$

1. Recognise the form. This is vertex form: $f(x) = -(x-1)^2 + 10$. Here $a = -1 < 0$, so the parabola opens downward — only a maximum exists.

2. Find the vertex. The squared term is zero when $x - 1 = 0$, i.e., $x = 1$. At that point, $-(x-1)^2 = 0$, so $f(1) = 0 + 10 = 10$.

   Since $-(x-1)^2 \le 0$ for all $x$, we have $f(x) \le 10$ for all $x$. The value $10$ is attained at $x = 1$.

3. Check for a minimum. As $x \to \pm\infty$, $-(x-1)^2 \to -\infty$, so $f(x) \to -\infty$. No lower bound — no minimum.

(iv) $g(x) = x^3 + 1$

1. Identify the type. This is a cubic function, not a quadratic. Cubics have no global maximum or minimum because they go to $+\infty$ in one direction and $-\infty$ in the other.

2. Check the limits. As $x \to \infty$, $x^3 \to \infty$, so $g(x) \to \infty$. As $x \to -\infty$, $x^3 \to -\infty$, so $g(x) \to -\infty$.

   The function takes every real value — it is strictly increasing (since $g'(x) = 3x^2 \ge 0$ and zero only at $x=0$). There is no highest or lowest value.

Method: Extrema of Vertex-Form and Odd-Degree Functions

This method applies whenever a function is presented as a perfect square (or its negative) plus a constant, or as an odd-degree polynomial like $x^3$ — the two commonest "spot the extremum" shapes tested together.

Steps

Step 1: Identify whether the expression is a shifted square or an odd-degree power.

A term like $(mx-c)^2$ (or $a(x-h)^2$) is always $\ge 0$, so adding/subtracting a constant simply shifts a definite minimum (or, with a leading minus sign, a definite maximum) up or down. An odd-degree term such as $x^3$ has no such bound — it runs to $-\infty$ on one side and $+\infty$ on the other.

Step 2: For a squared expression, set the inner bracket to zero.

If $f(x) = (mx+c)^2 + k$, the square is zero — and hence smallest — exactly when $mx + c = 0$, giving

Substituting this back gives the extreme value $k$ itself (or $-k$ for the negated form). Because the square term is unbounded above as $x \to \pm\infty$, this is the only extreme value the function has — a minimum if the square is added, a maximum if it is subtracted.

Step 3: For an odd-degree polynomial, check whether it is monotonic.

Differentiate: if $f'(x) \ge 0$ everywhere (zero only at isolated points), the function is increasing throughout $\mathbb{R}$, so it takes every real value and has neither a global maximum nor a global minimum — do not force a vertex formula onto it.

Step 4 (Applying to this problem): State the conclusion honestly for each part.

Report a genuine minimum/maximum only where one exists; for the odd-degree case, explicitly say "no maximum and no minimum" rather than leaving the question unanswered.

Common Mistakes

Mistake 1: Reading the vertex $x$-value directly off the coefficient instead of solving the bracket.

Why it's wrong: for $(2x-1)^2+3$, a student often assumes the vertex is at $x=1$ (mistaking the "$-1$" for a shift of $1$ unit) instead of correctly solving $2x-1=0$. Correct approach: always set the entire bracket equal to zero and solve for $x$ — here $2x-1=0 \Rightarrow x=\frac12$.

Mistake 2: Forcing a vertex formula onto a cubic.

Why it's wrong: $g(x)=x^3+1$ is not a quadratic, so it has no single turning point to plug into $x=-\frac{b}{2a}$ — attempting to do so produces a meaningless answer. Correct approach: recognise the odd degree first and check monotonicity via the derivative; if $g'(x)\ge 0$ everywhere, state plainly that no maximum or minimum exists.

Mistake 3: Completing the square incorrectly and getting the wrong minimum value.

Why it's wrong: in part (ii), rushing the algebra of $9x^2+12x+2$ can flip a sign and produce $x=\frac23$ instead of $x=-\frac23$. Correct approach: use $x=-\frac{b}{2a}$ directly as a check against the completed-square form before finalising the answer.