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Precalculus : Relations and Functions

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #51 : Functions

What is the domain of the function given by:

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

Possible Answers:

$\text{The set of all real numbers.}$

[-1,1]

$[-\frac{\pi}{2},\frac{\pi}{2}]$

[0,1]

$[0,\infty)$

Correct answer:

$\text{The set of all real numbers.}$

Explanation:

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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Example Question #52 : Functions

Find the domain of

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

Possible Answers:

$(-\infty,\sqrt[90]{31})$

$\text{The set of all real numbers}$

$(-\infty,\sqrt{31})$

$(-\sqrt[90]{31},\sqrt[90]{31})$

$(\sqrt[90]{31},\infty)$

Correct answer:

$\text{The set of all real numbers}$

Explanation:

Since $x^{90}+31 \neq 0$

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

of all real numbers.

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Example Question #1006 : Pre Calculus

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 f(x) ?

Possible Answers:

$\text{Domain}=(0, \infty)$

$\text{Range}= (-\infty, \infty)$

$\text{Domain}=(0,\infty)$

$\text{Range}= (0, \infty)$

$\text{Domain}=(1, \infty)$

$\text{Range}=(0,\infty)$

$\text{Domain}=(-\infty , \infty )$

$\text{Range}=(-\infty,\infty)$

$\text{Domain}=(0,\infty)$

$\text{Range}= (61.5, \infty )$

Correct answer:

$\text{Domain}=(0, \infty)$

$\text{Range}= (-\infty, \infty)$

Explanation:

Apart from the multiplication by 61.5 , 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 f(x) . 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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Example Question #53 : Functions

Find the domain and range of the given function.

y=x^2

Possible Answers:

Domain: All real numbers

Range: All real numbers

Domain: x>0

Range: y>0

Domain: x>0

Range: All real numbers

Domain: All real numbers

Range: $y\ge0$

Correct answer:

Domain: All real numbers

Range: $y\ge0$

Explanation:

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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Example Question #54 : Functions

Find the domain and range of the function.

$y=\sqrt{x}$

Possible Answers:

Domain: x>0

Range: All real numbers

Domain: All real numbers

Range: y>0

Domain: $x\ge0$

Range: $y\ge0$

Domain: All real numbers

Range: All real numbers

Correct answer:

Domain: $x\ge0$

Range: $y\ge0$

Explanation:

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 are never negative,

and the values for y are never negative,

Domain: $x\ge0$

Range: $y\ge0$

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Example Question #44 : Relations And Functions

What is the domain of $y=\sqrt{x-6}$

Possible Answers:

$[6,\infty)$

$(-6,\infty )$

$[-6,\infty )$

$(-\infty, \infty )$

$(6.\infty )$

Correct answer:

$[6,\infty)$

Explanation:

As long as the number under the square root sign is greater than or equal to , then the corresponding x-value is in the domain. So to figure out our domain, it is easiest to look at the equation and determine what is NOT in the domain. We do this by solving $\sqrt{x-6}=0$ and we get x=6 . We now look at values greater than and less than , and we can see that when x<6 , the number under the square root will be negative. When x>6 , the number will be greater than or equal to . Therefore, our domain is anything greater than or equal to 6, or $[6,\infty )$ .

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Example Question #45 : Relations And Functions

What is the range of y=x^2-4

Possible Answers:

$(-\infty ,-4]$

$(-4,\infty )$

$(4,\infty )$

$[-4,\infty )$

$[4,\infty )$

Correct answer:

$[-4,\infty )$

Explanation:

Because the only term in the equation containing an is squared, we know that its value will range from (when x=0 ) to $\infty$ (as approaches $\infty$ ). When x^2 is large, a constant such as does not matter, but when x^2 is at its smallest, it does. We can see that when x^2=0 , will be at its minimum of . This number gets bracket notation because there is an value such that y=-4 .

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Example Question #54 : Functions

What is the domain of the function?

$f(x) = e^{2x^{2}}$

Possible Answers:

$(-\infty, 1)$

$(-\infty, \infty)$

$(-\infty, 0)$

$(1, \infty)$

$(0, \infty)$

Correct answer:

$(-\infty, \infty)$

Explanation:

Any value can be inputed in the exponetial.

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Example Question #1014 : Pre Calculus

Find the domain of the function:

$\frac{1}{\sqrt{x+4}}$

Possible Answers:

$(-\infty,-4]$

$[-4,\infty )$

$(-\infty,\infty )$

$(-\infty,-4)$

$(-4,\infty )$

Correct answer:

$(-4,\infty )$

Explanation:

The square cannot house any negative term or can the denominator be zero. So the lower limit is since cannot be , but any value greater than it is ok. And the upper limit is infinity.

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Example Question #47 : Relations And Functions

What is the domain for the function?

$f(x) = \frac{\sqrt{2+3x}}{(x+4)(x-2)}$

Possible Answers:

$(-\infty,\infty)$

$(-4,-\frac{2}{3}]\cup[-\frac{2}{3},2 )\cup(2,\infty )$

$[-\frac{2}{3},2 )\cup(2,\infty )$

$(-4,-\frac{2}{3}]\cup[-\frac{2}{3},2 )$

$(-\infty,-4)\cup(-4,-\frac{2}{3}]\cup[-\frac{2}{3},2 )\cup(2,\infty )$

Correct answer:

$[-\frac{2}{3},2 )\cup(2,\infty )$

Explanation:

The denominator becomes when x = -4 or , so the function does not exist at these points. In numerator, must be at least $-\frac{2}{3}$ or greater to be real. So the function is continuous from $-\frac{2}{3}$ to and to any other value greater than .

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Precalculus : Relations and Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #51 : Functions

Example Question #52 : Functions

Example Question #1006 : Pre Calculus

Example Question #53 : Functions

Example Question #54 : Functions

Example Question #44 : Relations And Functions

Example Question #45 : Relations And Functions

Example Question #54 : Functions

Example Question #1014 : Pre Calculus

Example Question #47 : Relations And Functions

All Precalculus Resources

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