Functions

The functions are $f(x)=\sqrt{x-2}$ for $x\ge2$, and $g(x)=3x^2-6x+5$ for $0\le x\le4$, with real inputs. (i) Sketch y=f(x) and state its range. (ii) Find the inverse of f and its domain. (iii) Show that fg exists. (iv) Find fg(x) and its range. (v) With g now defined on all real numbers, describe fully a sequence of transformations taking…

Determine the range of each of the following functions. State, with reasons, whether the inverse function exists in each case. Then, find the rule and domain of the inverse function if it exists. (a) $h:x\mapsto\cos(x+1)$, $-1\leq x\leq-1+\dfrac{\pi}{2}$. (b) $q:x\mapsto e^x(x-1)$, $x\in\mathbb R$.

Complete the following table. | Row | Set notation | Interval notation | | --- | --- | --- | | 1 | Find the set notation | $[-3,5)$ | | 2 | $\{x\in\mathbb R:1<x\le10\}$ | Find the interval notation | | 3 | $\{y\in\mathbb R:y\ge-1\}$ | Find the interval notation | | 4 | Find the set notation | $(-\infty,-6)$ | | 5 | $\{w\in\mathbb R:w\ne2\}$ | Find the…

The function $f$ is defined by $f:x\mapsto3x^2+1$, $x\in\mathbb R$. Find (a) $f(3)$, (b) $f(5x)$, (c) $f(2y+1)$.

(a) $f(x)=e^{x+3}-a$ for real $x\le4$; $g(x)=2x+5$ for all real x. The parameter a is constant. (i) Find the inverse of f and its domain in terms of a. (ii) Given that the composite exists and $g^{-1}f(-3)=15$, find a without finding the composite rule. (b) $h(x)=2^x$ for $0<x\le2$, and $h(x)=6-x^2/2$ for $2<x\le3$, with $h(x)=h(x-3)$ for all real…

The function $f$ is defined by $f(x)=x^2-5x$, $x\in\mathbb R$. Find (a) $f(x+2)$, (b) $f(x^2)$, (c) $[f(x)]^2$.

State and justify which of the following is a function. (a) $f:x\mapsto x^2+1$, $x\in\mathbb R$. (b) $g:x\mapsto\pm\sqrt{x}$, $x\in\mathbb R$, $x\ge0$.

For all real x, $f(x)=2/[1+(5x-1)^2]$. (i) Find its range. (ii) Explain why it has no inverse. (iii) Restrict its domain to $x\ge k$: find the least k that gives an inverse, and find that inverse. (iv) Sketch the restricted f and its inverse together, clearly showing their relationship. (v) Show algebraically that a solution of…

Find the range of each of the following functions. (a) $f:x\mapsto x^2+4x+3$, $x\in\mathbb R$, $x\leq-2$. (b) $g:x\mapsto e^{2x}$, $x\in\mathbb R$, $x>0$. (c) $h:x\mapsto\sin x$, $x\in\mathbb R$, $-\pi\leq x\leq\pi$.

Determine the range of each of the following functions. (a) $f:x\mapsto\ln x-x$, $x\in\mathbb R$, $1<x\le e^2$. (b) $g:x\mapsto(x+2)^2+e^x$, $x\in\mathbb R$, $x>-4$.

(a) $f(x)=x^2+\lambda x+2$ for real $x\le2$, with constant lambda. (i) Find the values of lambda for which an inverse exists. (ii) With $\lambda=-4$, define the inverse, including its domain. (b) $g(x)=x^2+x+1$ for all real x, and $h(x)=x+4/x$ for real $x>0$. (i) Show hg exists. (ii) Find its range.

(a) If f is one-one, determine with justification whether $ff^{-1}$ exists. (b) $g(x)=\ln x$ for real $x\ge1$; $h(x)=x^2-2x+2$ for real $x>0$. (i) Given gh exists, define it including its domain and range. (ii) Sketch g, its inverse, and $g^{-1}g$ together. (iii) Find the inputs satisfying $g^{-1}g(x)=gg^{-1}(x)$.

It is given that $g(x)=\begin{cases}3,&0<x\leq2,\\5-x,&2<x\leq4,\end{cases}$ and that $g(x)=g(x+4)$ for all real values of $x$. (i) Find $g(3)$, $g(8)$ and $g(101)$. (ii) Sketch the graph of $y=g(x)$ for $-2<x\leq8$.

Define $f(x)=(x-1)/x$ for real $x\ne0,1$. (i) Define its inverse. (ii) Show the composite $f^2$ exists and equals the inverse. (iii) State $f^3(x)$. (iv) Given $f^{2014}g(x)=1-e^{-x}$ for $x>0$, find g(x).

For real $x\ne-1/2$, $f(x)=|(1-2x)/(2x+1)|$; for all real x, $g(x)=\tan^{-1}(2x)$. (i) Show f is not one-one. Find the least k such that restricting the domain to $[k,\infty)$ makes it one-one. On this restricted domain, (ii) define the inverse of f. (iii) Find the exact range of $gf^{-1}$.

It is given that $f(x)=\begin{cases}7-x^2&\text{for }0<x\leq2,\\2x-1&\text{for }2<x\leq4\end{cases}$ and that $f(x)=f(x+4)$ for all real values of $x$. (i) Evaluate $f(27)+f(45)$. (ii) Sketch the graph of $y=f(x)$ for $-7\leq x\leq10$.

For all real x, $f(x)=\cos x$. For $-1<x\le1$, $g(x)=\ln(1/(1+x))$. Define h by the same logarithmic rule for $-1<x<1$, but set h(1)=−1. (i) Show fg exists. (ii) Expand fg through the cubic term. (iii) State $h^{-1}(-1)$, then define the piecewise inverse as p(x) for x>a and c for x=−1, finding a,c and p.

It is given that $f(x)=\begin{cases}\cos^{-1}x&\text{for }0\leq x<1,\\\dfrac\pi2x-\dfrac\pi2&\text{for }1\leq x<2\end{cases}$ and that $f(x)=f(x+2)$ for all real values of $x$. (i) Sketch the graph given by $y=f(x)$ for $-3\leq x\leq4$. (ii) For $0\leq x\leq4$, solve the inequality $f(x)>\dfrac\pi3$, giving your answer in the exact form.

(i) Express $\sin x+\sqrt3\cos x$ as $R\sin(x+\alpha)$ with exact positive R and alpha. (ii) Define f by that expression on $\pi/6\le x\le k$. Find the largest exact k for which f has an inverse, define the inverse including its domain, and state the inputs where $ff^{-1}(x)=f^{-1}f(x)$.

The functions $f$ and $g$ are defined as $f(x)=1-x^2$, $x\in\mathbb R$, $-1\leq x\leq1$, $g(x)=\sqrt{x}+2$, $x\in\mathbb R$, $x\geq0$. (i) Show that the function $gf$ exists. (ii) Define the function $gf$ and sketch the graph of $gf$, stating the range of $gf$. (iii) Explain why $fg$ does not exist.

The function $f$ is defined by $f:x\mapsto e^{|2-x|}$, $x\in\mathbb R$, $x>a$, where $a$ is a constant. It is given that $a$ is the least possible value such that $f^{-1}$ exists.

A function $f$ is said to be a self-inverse function if $f(x)=f^{-1}(x)$. The function $f$ is defined by $f(x)=\dfrac{ax+b}{cx-2}$, $x\in\mathbb R$, $x\ne2/c$, where $a, b, c, d$ are non-zero constants.

Given that $f(x)=\begin{cases}x^2&0\le x<2,\\7-x&2\le x<3.\end{cases}$ Sketch the graph of $y=f(x)$ and state the range of the function.

(a) The curve $y=f(x)$ has horizontal asymptote $y=k$ and cuts the axes at $(a, 0)$ and $(0, b)$, where $a, b, k$ are non-zero constants. Given that $f^{-1}$ exists, state if possible the coordinates where the following curves cut the axes and their asymptotes: (i) $y=f(2x-3)$, (ii) $y=f^{-1}(x)$.

It is given that $f(x)=\begin{cases}3+x&0<x\le2,\\7-x&2<x\le4,\end{cases}$ and $f(x)=f(x+4)$ for all real $x$. (i) Find $f(1)$, $f(8)$ and $f(29)$. (ii) Sketch $y=f(x)$ for $-3<x\le8$.

Sketch the curve $y^2+ax^2=5$ for $x\ge0$, $y\ge0$, where $a$ is positive.

Functions $f:x\mapsto1+\dfrac2{x-1}$, $x\in\mathbb R$, $x<1$; $g:x\mapsto\ln x$, $x\in\mathbb R$, $0<x<1$.

(a) Functions $f:x\mapsto(x-2)(x-4)$ for $x\in\mathbb R$, $x<4$; $g:x\mapsto x/(x+3)$ for $x\in\mathbb R$, $x\ne-3$.

$f:x\mapsto(x-7)(x+1)$, $x>0$.

Functions $f:x\mapsto(x^2-6x-6)/(x+1)$, $x\in\mathbb R$, $x\ne-1$; $g:x\mapsto(ax+1)/(x+b)$, $x\in\mathbb R$, $x\ne-b$, where $a, b$ are constants.