Let $n \geq 2$ be a natural number and $f : [ 0,1 ] \rightarrow \mathbb { R }$ be the function defined by $$f ( x ) = \begin{cases} n ( 1 - 2 n x ) & \text { if } 0 \leq x \leq \frac { 1 } { 2 n } \\ 2 n ( 2 n x - 1 ) & \text { if } \frac { 1 } { 2 n } \leq x \leq \frac { 3 } { 4 n } \\ 4 n ( 1 - n x ) & \text { if } \frac { 3 } { 4 n } \leq x \leq \frac { 1 } { n } \\ \frac { n } { n - 1 } ( n x - 1 ) & \text { if } \frac { 1 } { n } \leq x \leq 1 \end{cases}$$ If $n$ is such that the area of the region bounded by the curves $x = 0 , x = 1 , y = 0$ and $y = f ( x )$ is 4, then the maximum value of the function $f$ is
Let $n \geq 2$ be a natural number and $f : [ 0,1 ] \rightarrow \mathbb { R }$ be the function defined by
$$f ( x ) = \begin{cases} n ( 1 - 2 n x ) & \text { if } 0 \leq x \leq \frac { 1 } { 2 n } \\ 2 n ( 2 n x - 1 ) & \text { if } \frac { 1 } { 2 n } \leq x \leq \frac { 3 } { 4 n } \\ 4 n ( 1 - n x ) & \text { if } \frac { 3 } { 4 n } \leq x \leq \frac { 1 } { n } \\ \frac { n } { n - 1 } ( n x - 1 ) & \text { if } \frac { 1 } { n } \leq x \leq 1 \end{cases}$$
If $n$ is such that the area of the region bounded by the curves $x = 0 , x = 1 , y = 0$ and $y = f ( x )$ is 4, then the maximum value of the function $f$ is