Relations and Functions - Pre-Calculus

Card 0 of 176

Question

What is the domain of the following function:

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

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 $\ge 0$

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Question

Which of the following expressions is not a function?

Answer

Recall that an expression is only a function if it passes the vertical line test. Test this by graphing each function and looking for one which fails the vertical line test. (The vertical line test consists of drawing a vertical line through the graph of an expression. If the vertical line crosses the graph of the expression more than once, the expression is not a function.)

Functions can only have one y value for every x value. The only choice that reflects this is:

$y=\pm\sqrt{x^2-7x+5}$

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Question

Suppose we have the relation $\small (a,b)$ on the set of real numbers $\mathbb{R}$ whenever $\small a<b$ . Which of the following is true.

Answer

The relation is not a function because $\small (2,3)$ and $\small (2,4)$ hold. If it were a function, $\small (2,b)$ would hold only for one $\small b$ . But we know it holds for $\small b=3,4$ because $\small 2<3$ and $\small 2<4$ . Thus, the relation $\small (a,b)$ on the set of real numbers $\small \mathbb{R}$ is not a function.

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Question

Consider a family consisting of a two parents, Juan and Oksana, and their daughters Adriana and Laksmi. A relation $\small (a,b)$ is true whenever $\small b$ is the child of $\small a$ . Which of the following is not true?

Answer

The statement

"Even if the two parents had only one daughter, the relation would not be a function."

is not true because if they had only one daughter, say Adriana, then the only relations that would exist would be (Juan, Adriana) and (Oksana,Adriana), which defines a function.

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Question

Which of the following relations is not a function?

Answer

The definition of a function requires that for each input (i.e. each value of ), there is only one output (i.e. one value of ). For , each value of corresponds to two values of (for example, when , both and are correct solutions). Therefore, this relation cannot be a function.

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Question

Given the set of ordered pairs, determine if the relation is a function

Answer

A relation is a function if no single x-value corresponds to more than one y-value.

Because the mapping from goes to and

the relation is NOT a function.

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Question

What equation is perpendicular to $y = \frac{5}{2}x+12$ and passes throgh ?

Answer

First find the reciprocal of the slope of the given function.
$m_2 = -\frac{1}{m_1}$

$m_2 = -\frac{2}{5}$

The perpendicular function is:

$y = -\frac{2}{5}x+c$

Now we must find the constant, , by using the given point that the perpendicular crosses.

$5 = -\frac{2}{5}(2)+c$

solve for :

$c = \frac{29}{5}$

$y = -\frac{2}{5}x + \frac{29}{5}$

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Question

Is the following relation of ordered pairs a function?

Answer

A set of ordered pairs is a function if it passes the vertical line test.

Because there are no more than one corresponding value for any given value, the relation of ordered pairs IS a function.

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Question

Find the range of the following function:

Answer

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

the range of f is :{7}.

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Question

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

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

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 :

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

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Question

What is the range of :

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:

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Question

Find the range of f(x) given below:

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 $[2,\infty)$

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Question

What is the range of :

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

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Question

Find the domain of the following function:

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

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.

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Question

What is the domain of the function given by:

$f(x)=\sqrt{1+cos^2(x)}$

Answer

cos(x) is definded for all reals. cos(x) is always between -1 and 1. Thus $1+cos^2(x)\ge0$ . The value inside the square is always positive. Therefore the domain is the set of all real numbers.

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Question

Find the domain of f(x) below

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

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: ${x\in \mathbb{R}: x e -1 }\cap {x\in \mathbb{R}: x\ge 7}={x\in \mathbb{R}: x\ge 7}$

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Question

The domain of the following function

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

is:

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 eq 2}=(1,2)\cup (2,\infty)$

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Question

Find the domain of

$\frac{x^{23}}{x^{90}+31}$

Answer

Since $x^{90}+31 eq 0$

for all real numbers, the denominator is never 0 .Therefore the domain is the set

of all real numbers.

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Question

Information about Nernst Equation:

http://physiologyweb.com/calculators/nernst\_potential\_calculator.html

The Nernst equation is very important in physiology, useful for measuring an ion's potential across cellular membranes. Suppose we are finding an ion's potential of a potassium ion at body temperature. Then the equation becomes:

$y = f(x) = 61.5 \log_{10} x$

Where is the ion's electrical potential in miniVolts and is a ratio of concentration.

What is the domain and range of ?

Answer

Apart from the multiplication by , this function is very similar to the function $f(x)=\log_{10}x$

The logarithmic function has domain x>0, meaning for every value x>0, the function $\log{ x}$ has an output (and for x = 0 or below, there are no values for log x)

The range indicates all the values that can be outputs of the . When you draw the graph of y = log(x), you can see that the function extends from -infinity (near x = 0), and then extends out infinitely in the positive x direction.

Therefore the Domain: $(0,\infty)$

and the is Range: $(-\infty, \infty)$

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Question

Find the domain and range of the given function.

Answer

The domain is the set of x-values for which the function is defined.

The range is the set of y-values for which the function is defined.

Because the values for x can be any number in the reals,

and the values for y are never negative,

Domain: All real numbers

Range: $y\ge0$

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