## MATH1901 Quizzes

Quiz 3: Functions
Question 1 Questions
What is the largest possible domain and the corresponding range of the following function?
$f\left(x\right)=ln\left(x+2\right)$
Exactly one option must be correct)
 a) Domain : $ℝ$, Range : $ℝ$ b) Domain : $\left(2,\infty \right)$, Range : $ℝ$ c) Domain : $\left(-2,\infty \right)$, Range : $\left[0,\infty \right)$ d) Domain : $\left[-2,\infty \right)$, Range : $\left[0,\infty \right)$ e) Domain : $\left(-2,\infty \right)$, Range : $ℝ$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is correct!
What is the largest possible domain and the corresponding range of the function
$f\left(x\right)=sin\left({e}^{x}\right)\phantom{\rule{1em}{0ex}}?$
Exactly one option must be correct)
 a) Domain : $\left(0,\infty \right)$, Range : $\left[-1,1\right]$ b) Domain : $ℝ$, Range : $\left[-1,1\right]$ c) Domain : $ℝ$, Range : $ℝ$ d) Domain : $\left(0,\infty \right)$, Range : $ℝ$ e) Domain : $\left[0,2\pi \right]$, Range : $\left[-1,1\right]$

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
If $f\left(x\right)={x}^{2}$ and $g\left(x\right)=x+1$ then the composite function $\left(f\circ g\right)\left(x\right)$ is equal to Exactly one option must be correct)
 a) ${x}^{2}+1$ b) $\left({x}^{2}+1\right)+1$ c) ${\left({x}^{2}+1\right)}^{2}$ d) ${\left(x+1\right)}^{2}$ e) $\sqrt{x+1}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Choice (e) is incorrect
If $f\left(x\right)=\sqrt{x}$, $g\left(x\right)=x+1$ and $h\left(x\right)={e}^{x}$ then $\left(\rightf\circ g\circ h\left)\right\left(x\right)$ is given by Exactly one option must be correct)
 a) $\sqrt{{e}^{x+1}}$ b) $\sqrt{{e}^{x}+1}$ c) ${e}^{\frac{x}{2}}+1$ d) ${e}^{\left(\sqrt{x}+1\right)}$ e) $\sqrt{{e}^{x}}+1$

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
The notation $f:A\to B$ means Exactly one option must be correct)
 a) the function $f$ takes all elements of set $A$ as inputs and produces all elements of set $B$ as outputs. b) the function $f$ takes all elements of set $A$ as inputs and the outputs will always be in set $B$. c) the function $f$ takes some subset of $A$ as inputs and the outputs will always be in set $B$. d) the domain of the function $f$ is $A$ and its range is $B$. e) none of the above.

Choice (a) is incorrect
$B$ is the codomain of $f$, so the range of $f$ is a subset of $B$. The range is not necessarily equal to $B$.
Choice (b) is correct!
$f:A\to B$’ means that the function is properly defined on all elements of $A$ and that its input values come from $A$. Its output values all lie in $B$ ( the target set or codomain). However $B$ need not be equal to the range. The range ( the set of values which $f$ actually maps to ) is a subset of $B$ and may be equal to $B$.
Choice (c) is incorrect
$A$ is equal to the domain of $f$.
Choice (d) is incorrect
$B$ is the codomain of $f$, not necessarily the range.
Choice (e) is incorrect
Consider the functions:
$f:ℝ\to ℝ;\phantom{\rule{1em}{0ex}}f\left(x\right)={x}^{4}+1$
$g:\left[0,\infty \right)\to ℝ;\phantom{\rule{1em}{0ex}}g\left(x\right)={x}^{4}+1$
$h:ℝ\to \left[1,\infty \right);\phantom{\rule{1em}{0ex}}h\left(x\right)={x}^{4}+1$
Check all statements which are true. (Zero or more options can be correct)
 a) $f$ is a surjective function. b) $g$ is a surjective function. c) $h$ is a surjective function. d) All three functions are surjective. e) $f$ and $h$ are surjective functions, and $g$ is not.

There is at least one mistake.
For example, choice (a) should be False.
$f$ is not surjective, since its range is $\left[1,\infty \right)$, not $ℝ$.
There is at least one mistake.
For example, choice (b) should be False.
$g$ is not surjective, since its range is $\left[1,\infty \right)$, not $ℝ$.
There is at least one mistake.
For example, choice (c) should be True.
There is at least one mistake.
For example, choice (d) should be False.
The only surjective function is $h$.
There is at least one mistake.
For example, choice (e) should be False.
The only surjective function is $h$.
Correct!
1. False $f$ is not surjective, since its range is $\left[1,\infty \right)$, not $ℝ$.
2. False $g$ is not surjective, since its range is $\left[1,\infty \right)$, not $ℝ$.
3. True
4. False The only surjective function is $h$.
5. False The only surjective function is $h$.
Consider the functions:
$f:ℝ\to ℝ;\phantom{\rule{1em}{0ex}}f\left(x\right)={e}^{x}$
$g:ℝ\to \left[1,\infty \right);\phantom{\rule{1em}{0ex}}g\left(x\right)={x}^{4}+1$
$h:\left[0,\infty \right)\to ℝ;\phantom{\rule{1em}{0ex}}h\left(x\right)={x}^{4}+1$
Which of the following statements are true? (Zero or more options can be correct)
 a) $f$ is an injective function. b) $f$ is surjective. c) $g$ is an injective function. d) $g$ is surjective. e) $h$ is an injective function. f) $h$ is surjective.

There is at least one mistake.
For example, choice (a) should be True.
If ${e}^{a}={e}^{b}$ then $a=b$, so $f$ is injective.
There is at least one mistake.
For example, choice (b) should be False.
The range of $f$ is $\left(0,\infty \right)$, so $f$ is not surjective.
There is at least one mistake.
For example, choice (c) should be False.
As $g\left(-1\right)=2=g\left(1\right)$, $g$ is not injective.
There is at least one mistake.
For example, choice (d) should be True.
The range of $g$ is $\left[1,\infty \right)$, so $g$ is surjective.
There is at least one mistake.
For example, choice (e) should be True.
As ${x}^{4}+1$ is an increasing function on $\left[0,\infty \right)$, this is an injective function.
There is at least one mistake.
For example, choice (f) should be False.
The range of $h$ is $\left[1,\infty \right)$ so $h$ is not surjective.
Correct!
1. True If ${e}^{a}={e}^{b}$ then $a=b$, so $f$ is injective.
2. False The range of $f$ is $\left(0,\infty \right)$, so $f$ is not surjective.
3. False As $g\left(-1\right)=2=g\left(1\right)$, $g$ is not injective.
4. True The range of $g$ is $\left[1,\infty \right)$, so $g$ is surjective.
5. True As ${x}^{4}+1$ is an increasing function on $\left[0,\infty \right)$, this is an injective function.
6. False The range of $h$ is $\left[1,\infty \right)$ so $h$ is not surjective.
Check all of the following functions which are injective. (Zero or more options can be correct)
 a) $f:\left(0,\infty \right)\to ℝ;f\left(x\right)=lnx$ b) $f:\left[0,\infty \right)\to ℝ;f\left(x\right)=coshx$ c) $f:ℝ\to ℝ;f\left(x\right)={x}^{3}+1$ d) $f:ℝ\to ℝ;f\left(x\right)=sinx$

There is at least one mistake.
For example, choice (a) should be True.
There is at least one mistake.
For example, choice (b) should be True.
There is at least one mistake.
For example, choice (c) should be True.
There is at least one mistake.
For example, choice (d) should be False.
Correct!
1. True
2. True
3. True
4. False
If $f:\left(0,\infty \right)\to ℝ;\phantom{\rule{1em}{0ex}}f\left(x\right)={x}^{2}+1$, and $g:ℝ\to \left(0,\infty \right);\phantom{\rule{1em}{0ex}}g\left(x\right)={e}^{x},$ find formulas for the composite function $f\circ g$ and the inverse function of $f$. Exactly one option must be correct)
 a) $\left(f\circ g\right)\left(x\right)={e}^{2x}+1$ and ${f}^{-1}\left(x\right)=\sqrt{x-1}$ b) $\left(f\circ g\right)\left(x\right)={e}^{2x}+1$ and ${f}^{-1}\left(x\right)=-\sqrt{x-1}$ c) $\left(f\circ g\right)\left(x\right)={e}^{2x}+1$ and ${f}^{-1}\left(x\right)=±\sqrt{x-1}$ d) $\left(f\circ g\right)\left(x\right)={e}^{{x}^{2}}+1$ and ${f}^{-1}\left(x\right)=\sqrt{x-1}$ e) $\left(f\circ g\right)\left(x\right)={e}^{{x}^{2}}+1$ and ${f}^{-1}\left(x\right)=-\sqrt{x-1}$

Choice (a) is correct!
Note that $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
Consider the function $f\left(x\right)=\sqrt{3+4x}$ on its natural domain. Find the inverse function ${f}^{-1}$, giving its domain and range. Exactly one option must be correct)
 a) ${f}^{-1}:ℝ\to ℝ$ and ${f}^{-1}\left(x\right)=\frac{{x}^{2}-3}{4}$ b) ${f}^{-1}:ℝ\to ℝ$ and ${f}^{-1}\left(x\right)=\frac{1}{\sqrt{3+4x}}$ c) ${f}^{-1}:\left[0,\infty \right)\to \left[-3∕4,\infty \right)$ and ${f}^{-1}\left(x\right)=\frac{{x}^{2}-3}{4}$ d) ${f}^{-1}:ℝ\to \left[-3∕4,\infty \right)$ and ${f}^{-1}\left(x\right)=\frac{{x}^{2}-3}{4}$ e) ${f}^{-1}:\left[0,\infty \right)\to ℝ$ and ${f}^{-1}\left(x\right)=\frac{{x}^{2}-3}{4}$ f) ${f}^{-1}:\left[0,\infty \right)\to ℝ$ and ${f}^{-1}\left(x\right)=\frac{1}{\sqrt{3+4x}}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Choice (e) is incorrect
Choice (f) is incorrect