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 $x=-2$


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