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Precalculus : Pre-Calculus

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Example Questions

Example Question #1021 : Pre Calculus

What is the domain of the function?

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

Possible Answers:

$(-\infty,\infty)$

$(0,\infty)$

Does not exist anywhere.

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

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 #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 #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 #1 : Determine If A Relation Is A Function

Which of the following expressions is not a function?

Possible Answers:

y=5x+99

y=2^x+343

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

y=6x^2-54x+76

y=sinx

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 #2 : Determine If A Relation Is A Function

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

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

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

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

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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Example Question #3 : Determine If A Relation Is A Function

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?

Possible Answers:

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

(Adriana,Laksmi) does not hold because Laksmi is not Adriana's child and is a boy.

(Laksmi,Adriana) does not hold because Adriana is not Laksmi's child.

The relation is not a function because (Laksmi,Juan) and (Laksmi,Oksana) both hold.

If the two parents had only one daughter, the relation would be a function.

Correct answer:

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

Explanation:

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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Example Question #3 : Determine If A Relation Is A Function

Which of the following relations is not a function?

Possible Answers:

y=2

x = y^2

$y = x^{\frac{3}{2}}$

$y = \sqrt{x}$

Correct answer:

x = y^2

Explanation:

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 x = y^2 , each value of corresponds to two values of (for example, when x = 4 , both y=2 and y = -2 are correct solutions). Therefore, this relation cannot be a function.

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Example Question #4 : Determine If A Relation Is A Function

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

(0,2)

(1,4)

(2,6)

(3,8)

(2,5)

Possible Answers:

Yes

Cannot be determined

Correct answer:

Explanation:

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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Example Question #1 : Determine If A Relation Is A Function

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

Possible Answers:

$y = -\frac{5}{2}x + \frac{22}{5}$

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

$y = -\frac{4}{5}x + \frac{26}{5}$

$y = \frac{2}{5}x + \frac{21}{5}$

$y = -\frac{2}{5}x + \frac{27}{5}$

Correct answer:

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

Explanation:

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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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1021 : Pre Calculus

Example Question #54 : Relations And Functions

Example Question #54 : Relations And Functions

Example Question #1024 : Pre Calculus

Example Question #1 : Determine If A Relation Is A Function

Example Question #2 : Determine If A Relation Is A Function

Example Question #3 : Determine If A Relation Is A Function

Example Question #3 : Determine If A Relation Is A Function

Example Question #4 : Determine If A Relation Is A Function

Example Question #1 : Determine If A Relation Is A Function

All Precalculus Resources

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