Piecewise Functions and Integration
DAILY_CHALLENGE
Grade None
Question:
Let $n\geq2$ be a natural number and $f:[0,1]\to\mathbb{R}$ be the function defined by
$$f(x)=\begin{cases}n(1-2nx) & \text{if } 0\leq x\leq\dfrac{1}{2n},\\ 2n(2nx-1) & \text{if } \dfrac{1}{2n}\leq x\leq\dfrac{3}{4n},\\ 4n(1-nx) & \text{if } \dfrac{3}{4n}\leq x\leq\dfrac{1}{n},\\ \dfrac{n}{n-1}(nx-1) & \text{if } \dfrac{1}{n}\leq x\leq1.\end{cases}$$
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 ___.
Step-by-Step Solution
Key Concept: Integrate each piecewise segment; area = n/2, so n=8; max of f equals n throughout
Compute $\int_0^1 f(x)\,dx$ piece by piece:
$\int_0^{1/(2n)} n(1-2nx)dx = n[x-nx^2]_0^{1/(2n)} = \dfrac{1}{4}$.
$\int_{1/(2n)}^{3/(4n)} 2n(2nx-1)dx$: sub $u=2nx-1$, range $[0,1/2]$: $=\int_0^{1/2}u\,du=\dfrac{1}{8}$.
$\int_{3/(4n)}^{1/n}4n(1-nx)dx=\dfrac{1}{8}$.
$\int_{1/n}^1\dfrac{n}{n-1}(nx-1)dx=\dfrac{n}{n-1}\cdot\dfrac{(n-1)^2}{2n}=\dfrac{n-1}{2}$.
Total area $=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{n-1}{2}=\dfrac{1}{2}+\dfrac{n-1}{2}=\dfrac{n}{2}=4\Rightarrow n=8$.
Maximum of $f$: checking all pieces at $n=8$: $f(0)=8$, peak of 3rd piece at $x=3/(4\cdot8)=3/32$: $f=8$. $f(1)=8$.
Max value $=n=8$.
Correct Answer: 8