Domain and Range - Pre-Calculus

Card 1 of 108

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

What is the domain of the following function:

$f(x)=\frac{1}{1+\sqrt x}$

Tap to reveal answer

Answer

Note that in the denominator, we need to have $x\ge 0$ to make the square root of x defined. In this case $1+\sqrt x$ is never zero. Hence we have no issue when dividing by this number. Therefore the domain is the set of real numbers that are $\ge 0$

← Didn't Know|Knew It →

Question

What is the domain of the following function:

$f(x)=\frac{1}{1+\sqrt x}$

Tap to reveal answer

Answer

Note that in the denominator, we need to have $x\ge 0$ to make the square root of x defined. In this case $1+\sqrt x$ is never zero. Hence we have no issue when dividing by this number. Therefore the domain is the set of real numbers that are $\ge 0$

← Didn't Know|Knew It →

Question

What is the domain of the following function:

$f(x)=\frac{1}{1+\sqrt x}$

Tap to reveal answer

Answer

Note that in the denominator, we need to have $x\ge 0$ to make the square root of x defined. In this case $1+\sqrt x$ is never zero. Hence we have no issue when dividing by this number. Therefore the domain is the set of real numbers that are $\ge 0$

← Didn't Know|Knew It →

Question

The domain of the following function

$f(x)=\frac{1}{\sqrt{x-1}-1}$

is:

Tap to reveal answer

Answer

$\sqrt{x-1}$ is defined when $x\ge 1$ . Since we do not want $\sqrt{x-1}$ to be 0 in the denominator we must have $x\ >1$ . $\sqrt{x-1}-1=0$ when x=2.

Thus we need to exclude 2 also. Therefore the domain is:

$$\{x\in \mathbb{R} : x>1\cap x\neq 2\}=(1,2)\cup (2,\infty)$

← Didn't Know|Knew It →

Question

Find the domain of f(x) below

$\frac{\sqrt{x-7}}{x^3+1}$

Tap to reveal answer

Answer

We have We have $x^2-x+1\ >0$ for all real numbers. when . The denominator is undefined when .

The nominator is defined if $x\ge 7$ .

The domain is: $\{x\in \mathbb{R}: x\ne -1 \}\cap \{x\in \mathbb{R}: x\ge 7\}=\{x\in \mathbb{R}: x\ge 7\}$

← Didn't Know|Knew It →

Question

The domain of the following function

$f(x)=\frac{1}{\sqrt{x-1}-1}$

is:

Tap to reveal answer

Answer

$\sqrt{x-1}$ is defined when $x\ge 1$ . Since we do not want $\sqrt{x-1}$ to be 0 in the denominator we must have $x\ >1$ . $\sqrt{x-1}-1=0$ when x=2.

Thus we need to exclude 2 also. Therefore the domain is:

$$\{x\in \mathbb{R} : x>1\cap x\neq 2\}=(1,2)\cup (2,\infty)$

← Didn't Know|Knew It →

Question

Find the domain of f(x) below

$\frac{\sqrt{x-7}}{x^3+1}$

Tap to reveal answer

Answer

We have We have $x^2-x+1\ >0$ for all real numbers. when . The denominator is undefined when .

The nominator is defined if $x\ge 7$ .

The domain is: $\{x\in \mathbb{R}: x\ne -1 \}\cap \{x\in \mathbb{R}: x\ge 7\}=\{x\in \mathbb{R}: x\ge 7\}$

← Didn't Know|Knew It →

Question

The domain of the following function

$f(x)=\frac{1}{\sqrt{x-1}-1}$

is:

Tap to reveal answer

Answer

$\sqrt{x-1}$ is defined when $x\ge 1$ . Since we do not want $\sqrt{x-1}$ to be 0 in the denominator we must have $x\ >1$ . $\sqrt{x-1}-1=0$ when x=2.

Thus we need to exclude 2 also. Therefore the domain is:

$$\{x\in \mathbb{R} : x>1\cap x\neq 2\}=(1,2)\cup (2,\infty)$

← Didn't Know|Knew It →

Question

Find the domain of f(x) below

$\frac{\sqrt{x-7}}{x^3+1}$

Tap to reveal answer

Answer

We have We have $x^2-x+1\ >0$ for all real numbers. when . The denominator is undefined when .

The nominator is defined if $x\ge 7$ .

The domain is: $\{x\in \mathbb{R}: x\ne -1 \}\cap \{x\in \mathbb{R}: x\ge 7\}=\{x\in \mathbb{R}: x\ge 7\}$

← Didn't Know|Knew It →

Question

Find the domain of f(x) below

$\frac{\sqrt{x-7}}{x^3+1}$

Tap to reveal answer

Answer

We have We have $x^2-x+1\ >0$ for all real numbers. when . The denominator is undefined when .

The nominator is defined if $x\ge 7$ .

The domain is: $\{x\in \mathbb{R}: x\ne -1 \}\cap \{x\in \mathbb{R}: x\ge 7\}=\{x\in \mathbb{R}: x\ge 7\}$

← Didn't Know|Knew It →

Question

The domain of the following function

$f(x)=\frac{1}{\sqrt{x-1}-1}$

is:

Tap to reveal answer

Answer

$\sqrt{x-1}$ is defined when $x\ge 1$ . Since we do not want $\sqrt{x-1}$ to be 0 in the denominator we must have $x\ >1$ . $\sqrt{x-1}-1=0$ when x=2.

Thus we need to exclude 2 also. Therefore the domain is:

$$\{x\in \mathbb{R} : x>1\cap x\neq 2\}=(1,2)\cup (2,\infty)$

← Didn't Know|Knew It →

Question

Find the domain of the following function f(x) given below:

$f(x)=\sqrt{ x^3-1}$

Tap to reveal answer

Answer

. Since $(x^2-x+1)\ >0$ for all real numbers. To make the square root positive we need to have $x-1\ge 0$ .

Therefore the domain is :

$\{x\in\mathbb{R}: x\ge 1\}=[1,\infty)$

← Didn't Know|Knew It →

Question

Find the domain of the following function f(x) given below:

$f(x)=\sqrt{ x^3-1}$

Tap to reveal answer

Answer

. Since $(x^2-x+1)\ >0$ for all real numbers. To make the square root positive we need to have $x-1\ge 0$ .

Therefore the domain is :

$\{x\in\mathbb{R}: x\ge 1\}=[1,\infty)$

← Didn't Know|Knew It →

Question

Find the domain of the following function f(x) given below:

$f(x)=\sqrt{ x^3-1}$

Tap to reveal answer

Answer

. Since $(x^2-x+1)\ >0$ for all real numbers. To make the square root positive we need to have $x-1\ge 0$ .

Therefore the domain is :

$\{x\in\mathbb{R}: x\ge 1\}=[1,\infty)$

← Didn't Know|Knew It →

Question

Find the domain of the following function f(x) given below:

$f(x)=\sqrt{ x^3-1}$

Tap to reveal answer

Answer

. Since $(x^2-x+1)\ >0$ for all real numbers. To make the square root positive we need to have $x-1\ge 0$ .

Therefore the domain is :

$\{x\in\mathbb{R}: x\ge 1\}=[1,\infty)$

← Didn't Know|Knew It →

Question

Find the range of the following function:

Tap to reveal answer

Answer

Every element of the domain has as image 7.This means that the function is constant . Therefore,

the range of f is :{7}.

← Didn't Know|Knew It →

Question

What is the range of :

Tap to reveal answer

Answer

We know that $-1\le cos(x)\le 1$ . So $-2\le 2cos(x)\le 2$ .

Therefore:

$1-2\le 1+2cos(x)\le 1+2$ .

This gives:

$-1\le 1+2cos(x)\le 3$ .

Therefore the range is:

← Didn't Know|Knew It →

Question

Find the range of f(x) given below:

Tap to reveal answer

Answer

Note that: we can write f(x) as :

.

Since,

$(x+1)^2\ge 0 $ for all reals.$

Therefore,

$(x+1)^2+2\ge 2.$

So the range is $[2,\infty)$

← Didn't Know|Knew It →

Question

What is the range of :

Tap to reveal answer

Answer

We have $-1\le cos(3x)\le 1$ .

Adding 7 to both sides we have:

$6\le cos(3x)\le 8$ .

Therefore $6\le f(x)\le 8$ .

This means that the range of f is

← Didn't Know|Knew It →

Question

Find the domain of the following function:

$f(x)=\sqrt{121-x}$

Tap to reveal answer

Answer

The part inside the square root must be positive. This means that we must be $\ge 0$ . Thus

$121-x\ge0$ . Adding -121 to both sides gives $-x\ge -121$ . Finally multiplying both sides by (-1) give:

$x\le 121$ with x reals. This gives the answer.

Note: When we divide by a negative we need to flip our sign.

← Didn't Know|Knew It →

0

Knew It