hi!
Attached is a function qns......now studying for a test and come across problems which i have difficulty completing.... thanks for ur help in advance!
A function $\displaystyle f$ is defined by:
$\displaystyle f:x\mapsto ax+b,\ x \in \mathbb{R}$,
where $\displaystyle a$ and $\displaystyle b$ are positive constants.
(a) i. Find $\displaystyle f^3(x),$ (where $\displaystyle f^3(x)=fff(x)$).
$\displaystyle f^2(x) = f(f(f(x))) = f(f(ax+b))$
$\displaystyle =f(a(ax+b)+b)$$\displaystyle =a(a(ax+b)+b)+b$$\displaystyle =a^3x+a^2b+ab+b$(ii) Given that $\displaystyle f^3(x)=64x+21$, find the values of $\displaystyle a$ and $\displaystyle b$.
From part (i) we know that the coefficient of $\displaystyle x$ is $\displaystyle a^3$, so we have $\displaystyle a^3=64$. or $\displaystyle a=4$.
We also know that the constant term is $\displaystyle a^2b+ab+b=21$, which when we substitute in the known value of $\displaystyle a$ gives us $\displaystyle b=1$.
RonL
(other parts to follow in another post if someone else does not provide their solutions first).
(iii) With the values of $\displaystyle a$ and $\displaystyle b$ found in part (ii) determin $\displaystyle f^n(x)$, leaving your answer in the form $\displaystyle p^nx+\frac{p^n-1}{q}$ where $\displaystyle p$ and $\displaystyle q$ are integers to be found.
From the way that part (i) went I will assume that:
$\displaystyle
f^n(x)=a^nx + a^{n-1}b + \dots +ab + b
$
Putting in $\displaystyle a=4$ and $\displaystyle b=1$:
$\displaystyle f^n(x)=4^n x + 4^{n+1}+\cdots + 4 + 1$
Which simplifies to:
$\displaystyle
f^n(x)=4^nx + \frac{4^n-1}{3}$
RonL
Another function $\displaystyle g$ is defined by $\displaystyle g: x\mapsto e^x,\ x \in \mathbb{R},\ x<0$.
(b) (i) Find in terms of $\displaystyle a$ and $\displaystyle b$ the range of $\displaystyle fg$.
With the definition of $\displaystyle g$ the range of $\displaystyle g$ is the open interval $\displaystyle (0,1)$, which $\displaystyle f$ will map to the open interval $\displaystyle (b, (b+a))$
(ii) If $\displaystyle h(x)=[g(x)]^2$, determine, with a reason, whether $\displaystyle h^{-1}$ exists.
With the give devinition of $\displaystyle g(x)$ we have $\displaystyle h(x)=e^{2x}$, and the range of $\displaystyle h$ is the open interval $\displaystyle (0,1)$. But $\displaystyle \ln$ is defined everywhere on this interval and single valued, so if:
$\displaystyle h(x)=y$,
then $\displaystyle x=\ln(y)/2$, and is the only such $\displaystyle x$ with image under $\displaystyle h$ of $\displaystyle y$
RonL