Let f:(1,)R, f(x)=x-1x+1.

Question

Let $f:(1,\infty)	o\mathbb{R}$, $f(x)=\dfrac{x-1}{x+1}$. Let $f^{i+1}(x)=f(f^i(x))$, $i=1,\dots,25$, $f^1=f$. Let $g(x)+f^{26}(x)=0$, $x\in(1,\infty)$. If the area of the region bounded by $y=g(x)$, $2y=2x-3$, $y=0$, $x=4$ is $A$, then $A$ is:

Accepted Answer

Computing the iterates: $f^2(x)=-\dfrac1x$, $f^3(x)=\dfrac{1+x}{1-x}$, $f^4(x)=x$ --- period 4. Since $25\equiv1\pmod4$, $f^{25}=f^1$, so $f^{26}=f^2(x)=-\dfrac1x$, giving $g(x)=\dfrac1x$.

The bounded region (slant line $y=x-\tfrac32$, curve $y=1/x$, line $x=4$, $x$-axis) has vertices $A(2,\tfrac12)$, $B(4,\tfrac14)$, $C(4,0)$, $D(\tfrac32,0)$. Splitting into triangle $AMD$ plus area under the hyperbola from $x=2$ to $4$:

$$\text{Area} = \frac12\cdot(2-1.5)\cdot\frac12 + \int_2^4\frac1x\,dx = \frac18+\ln2.$$

Solutions · 1