The curve $y=\mathrm{f}(x)$ cuts the axes at $(a,0)$ and $(0,b)$. It is given that ${{\mathrm{f}}^{-

Question

The curve $y=\mathrm{f}(x)$ cuts the axes at $(a,0)$ and $(0,b)$. It is given that ${{\mathrm{f}}^{-1}}(x)$ exists. State, if it is possible to do so, the coordinates of the points where the following curves cut the axes. $y=\mathrm{f}(2x)$ $y=\mathrm{f}(x-1)$ $y=\mathrm{f}(2x-1)$ $y={{\mathrm{f}}^{-1}}(x)$ [4] The function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto 1-\frac{1}{1-x}$, where $x\in \mathbb{R}$, $x
e a$. State the value of $a$ and explain why this value has to be excluded from the domain of $\mathrm{g}$. [2] Find ${{\mathrm{g}}^{2}}(x)$ and ${{\mathrm{g}}^{-1}}(x)$, giving your answers in simplified form. [4] Find the values of $b$ such that ${{\mathrm{g}}^{2}}(b)={{\mathrm{g}}^{-1}}(b)$. [2]

Accepted Answer

Transformations: (i) $(0,b)$, $(\frac a2,0)$; (ii) $(a+1,0)$; (iii) $(\frac12,b)$, $(\frac{a+1}{2},0)$; (iv) $(0,a)$, $(b,0)$. For $g$: $a=1$, $g^2(x)=x$, $g^{-1}(x)=1-\frac1{1-x}$, $b=0\text{ or }2$

N2017 P2 Q3

N2017 P2 Q3

Video Solution:

Default solution