Divided we fall

Algebra Level 1

What is the maximum value of $f(x)$ , defined in $0 \leq x \leq 4$ as $f(x) = \begin{cases} 2x & \text{ for } 0 \leq x \leq 2, \\ 8-2x & \text{ for } 2 < x \leq 4. \\ \end{cases}$

2 5 4 3

1 solution

Arron Kau Staff
May 13, 2014

Since $f(x)=2x$ is an increasing function in $0 \leq x \leq 2$ , it is maximized at the right end of the domain. Thus, $f(x) \leq f(2)=4$ , which implies that the maximum value is $4$ in the domain $0 \leq x \leq 2$ .

Since $f(x)=8-2x$ is a decreasing function in $2 < x \leq 4$ , it is maximized at the left end of the domain. Thus, $f(x) < f(2)=4$ , which implies that the maximum value is $< 4$ in the domain $2 \leq x \leq 4$ .

Therefore, the maximum value of the given function is $4$ .

Divided we fall

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