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Precalculus : Relations and Functions

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #51 : Relations And Functions

What is the domain of the function below?

$\frac{1}{x^2 + 3x + 2}$

Possible Answers:

$(-\infty, -2)\cup(-2,-1)\cup(-1,\infty )$

$(-\infty,\infty )$

$(-\infty, -3)\cup(-3,\infty )$

$(-\infty,1)\cup(1,2)\cup(2,\infty )$

$(-\infty, -2]\cup[-2,-1]\cup[-1,\infty )$

Correct answer:

$(-\infty, -2)\cup(-2,-1)\cup(-1,\infty )$

Explanation:

The denomiator factors out to:

(x+2)(x+1)

The denominator becomes zero when x = -1,-2 . But the function can exist at any other value.

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Example Question #62 : Functions

What is the domain of the function below?

$\sqrt{3+5x}$

Possible Answers:

$[\frac{3}{5},\infty )$

$(-\infty,3 )$

$[-3,\infty )$

$[-\frac{3}{5},\infty )$

Correct answer:

$[-\frac{3}{5},\infty )$

Explanation:

Cannot have a negative inside the square root. The value of has to be $-\frac{3}{5}$ for the inside of the square root to be at least . This is the lower bound of the domain. Any value of greater than $-\frac{3}{5}$ exists.

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Example Question #1016 : Pre Calculus

f(x) = ln((x+2)(x-3))

Possible Answers:

$(-\infty,\infty )$

$(-\infty,-2 )\cup(-2,3 )\cup(3,\infty )$

(-2,3)

$(3,\infty)$

$(-\infty,-2 )\cup(3,\infty )$

Correct answer:

$(-\infty,-2 )\cup(3,\infty )$

Explanation:

The natural log function does not exist if the inside value is negatuve or zero. The points where the inside becomes negative are x = 3 or . If is greater than , both terms, (x+2) and (x-3) , are positive. If is less than , both terms are negative and multiply to become positive. If the value is between and , only one term will be negative and result in a $ln(-\#)$ , which does not exist.

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Example Question #64 : Functions

What is the domain of the function?

f(x) = ln(3+x)

Possible Answers:

$[-3,\infty)$

$(\frac{1}{3},\infty)$

$[\frac{1}{3},\infty)$

$(-\infty,\infty)$

$(-3,\infty)$

Correct answer:

$(-3,\infty)$

Explanation:

The value inside a natural log function cannot be negative or . At x = -3 , the inside is and any value less than {-3} cannot be included, because result will be a negative number inside the natural log.

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Example Question #51 : Relations And Functions

What is the domain of the function?

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

Possible Answers:

$(0,\infty)$

Does not exist anywhere.

$(-\infty,\infty)$

$(\frac{1}{2},\infty)$

Correct answer:

$(-\infty,\infty)$

Explanation:

Exponentials cannot have negatives on the inside. However, the expoential will convert any value into a positive value.

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Example Question #52 : Relations And Functions

What is the domain of the function?

$f(x) = \frac{ln((x-2)^2)}{(3x)^{1/3}}$

Possible Answers:

$(-\infty,2)\cup(2,\infty)$

$(-\infty,0)\cup(2,\infty)$

$(-\infty,0)\cup(0,2)\cup(2,\infty)$

$(-\infty,0)\cup(0,\infty)$

$(-\infty,\infty)$

Correct answer:

$(-\infty,0)\cup(0,2)\cup(2,\infty)$

Explanation:

Looking at the denominator, the function cannot exist at x = 0 . The natural log function cannot have a or negative inside. Since the value is raised to the power of , any negative value will be convert to a positive value. However, the function will not exist if the inside of the natural is , where x = 2 . f(x) will exist any where else.

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Example Question #54 : Relations And Functions

What is the domain of the function?

$f(x) = \frac{ln(x+7)}{(x-3)}$

Possible Answers:

$(-7,3)\cup(3,\infty)$

$(-\infty,-7)\cup(-7,3)\cup(3,\infty)$

$(-\infty,\infty)$

$(3,\infty)$

Correct answer:

$(-7,3)\cup(3,\infty)$

Explanation:

The denominator becomes where x = 3 , and the inside of the natural log also becomes at x = -7 . The function will not exist at these two points. The value cannot be less than x = -7 , becuase that will leave a negative value inside in the natural log.

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Example Question #1024 : Pre Calculus

Find the domain of the function.
$f(x) = \frac{x^2+6x+9}{x^2+5x+6}$

Possible Answers:

$(-\infty,-3)\cup(-3,-2 )\cup(-2,\infty )$

$(-\infty,\infty)$

$(-\infty,-3)\cup(-3,\infty)$

$(-\infty,-2)\cup(-2,\infty)$

Correct answer:

$(-\infty,-3)\cup(-3,-2 )\cup(-2,\infty )$

Explanation:

Simplify:

$f(x) = \frac{x^2+6x+9}{x^2+5x+6} = \frac{(x+3)^2}{(x+3)(x+2)}$ $=\frac{x+3}{x+2}$

Even though the (x+3) cancels out from the numerator and denominator, there is still a hole where the function discontinues at x = -3 . The function also does not exist at x =-2 , where the denominator becomes .

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Example Question #52 : Relations And Functions

Which of the following expressions is not a function?

Possible Answers:

y=sinx

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

y=2^x+343

y=5x+99

y=6x^2-54x+76

Correct answer:

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

Explanation:

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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Example Question #53 : Relations And Functions

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.

Possible Answers:

The relation is a function because $\small (2,4)$ holds and $\small (3,4)$ also holds.

The relation is not a function because $\small (2,3)$ and $\small (2,4)$ both hold.

The relation is a function because every relation is a function, since that's how relations are defined.

The relation is a function because for every $\small a$ , there is only one $\small b$ such that $\small (a,b)$ holds.

The relation is not a function because $\small (0,1)$ holds but $\small (1,0)$ does not.

Correct answer:

The relation is not a function because $\small (2,3)$ and $\small (2,4)$ both hold.

Explanation:

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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Precalculus : Relations and Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #51 : Relations And Functions

Example Question #62 : Functions

Example Question #1016 : Pre Calculus

Example Question #64 : Functions

Example Question #51 : Relations And Functions

Example Question #52 : Relations And Functions

Example Question #54 : Relations And Functions

Example Question #1024 : Pre Calculus

Example Question #52 : Relations And Functions

Example Question #53 : Relations And Functions

All Precalculus Resources

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