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Precalculus : Functions

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Example Questions

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 #1012 : Pre Calculus

What is the domain for the function?

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

Possible Answers:

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

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

$(-\infty,\infty)$

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

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

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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Example Question #51 : Relations And Functions

What is the domain of the function below?

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

Possible Answers:

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

$(-\infty,\infty )$

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

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

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

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 #51 : Relations And Functions

What is the domain of the function below?

$\sqrt{3+5x}$

Possible Answers:

$[-3,\infty )$

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

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

$(-\infty,3 )$

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 #52 : Relations And Functions

What is the domain of the function?

f(x) = ln(3+x)

Possible Answers:

$(-3,\infty)$

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

$[-3,\infty)$

$(-\infty,\infty)$

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

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 #61 : Functions

What is the domain of the function?

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

Possible Answers:

Does not exist anywhere.

$(0,\infty)$

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

$(-\infty,\infty)$

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 #54 : Relations And Functions

What is the domain of the function?

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

Possible Answers:

$(-\infty,\infty)$

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

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

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

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

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 #55 : Relations And Functions

What is the domain of the function?

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

Possible Answers:

$(-\infty,\infty)$

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

$(3,\infty)$

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

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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Precalculus : Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1014 : Pre Calculus

Example Question #1012 : Pre Calculus

Example Question #51 : Relations And Functions

Example Question #51 : Relations And Functions

Example Question #1016 : Pre Calculus

Example Question #52 : Relations And Functions

Example Question #61 : Functions

Example Question #54 : Relations And Functions

Example Question #55 : Relations And Functions

Example Question #1024 : Pre Calculus

All Precalculus Resources

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