# Maximum and Minimum of Quadratics by Domain

### Example 1

Find the maximum and minimum values of $y=x^2$, for $1 \le x \le 2$.

From the graph, its maximum value is $4$, when $x=2$, and its minimum value is $1$, when $x=1$.

### Example 2

Find the maximum and minimum values of $y=x^2$, for $-2 \le x \le -1$.

From the graph, its maximum value is $4$, when $x=-2$, and its minimum value is $1$, when $x=-1$.

### Example 3

Find the maximum and minimum values of $y=x^2$, for $-1 \le x \le 2$.

From the graph, its maximum value is $4$, when $x=2$, and its minimum value is $0$, when $x=0$.

### Example 4

Find the maximum and minimum values of $y=x^2$, for $-2 \le x \le 1$.

From the graph, its maximum value is $4$, when $x=-2$, and its minimum value is $0$, when $x=0$.

### Example 5

Consider the function $3x^4+4x^3-12x^2-3$ for $-3 \le x \le 2$.

(a)   Find the x values of the stationary points.

\begin{align} \displaystyle f^{\prime}(x) &= 12x^3+12x^2-24x \\ 12x^3+12x^2-24x &= 0 \\ x^3+x^2-2x &= 0 \\ x(x^2+x-2)x &= 0 \\ x(x+2)(x-1) &=1 \\ \therefore x &= 0 \text{ or } x = x=-2 \text{ or } x=1 \end{align}

(b)   Find the values of $y$ of the stationary points and endpoints.

For stationary points;
\begin{align} \displaystyle f(-2) &= 3 \times (-2)^4 + 4 \times (-2)^3-12 \times (-2)^2-3 = -35 \\ f(0) &= 3 \times 0^4 + 4 \times 0^3-12 \times 0^2-3 = -3 \\ f(1) &= 3 \times 1^4 + 4 \times 1^3-12 \times 1^2-3 = -8 \\ \end{align}
For endpoints;
\begin{align} \displaystyle f(-3) &= 3 \times (-3)^4 + 4 \times (-3)^3-12 \times (-3)^2-3 = 24 \\ f(2) &= 3 \times 2^4 + 4 \times 2^3-12 \times 2^2-3 = 29 \end{align}

(c)   Find the maximum value.

$24$ at $x=-3$

(d)   Find the minimum value.

$-35$ at $x=-2$