Functions - Pre-Calculus

Card 0 of 644

Question

Which of the following is a point on the following function?

$y=\left | x^2-5x-56\right |+27$

Answer

One way to approach this problem would be to plug in each answer and see what works. However, I would be a little more strategic and eliminate any options that don't make sense.

Our y value will never be negative, so eliminate any options with a negative y-value.

Try (0,0) really quick, since it's really easy

$y=\left | x^2-5x-56\right |+27 \rightarrow 0=\left|0^2-5(0)-56\right|+27=83 \rightarrow 0 eq 83$

The only point that makes sense is (5,83), therefore it is the correct answer

$y=\left | x^2-5x-56\right |+27 \rightarrow 83=\left|5^2-5(5)-56\right|+27=83 \rightarrow 83=83$

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Question

Evaluate: $\left | x-3\right |<15$

Answer

Cancel the absolute value sign by separating the function $\left | x-3\right |<15$ into its positive and negative counterparts.

Evaluate the first scenario.

Evaluate the second scenario.

The correct answer is:

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Question

If $y=\left |x^3-5x^2+5 \right |$ , then what is the value of when ?

Answer

We evaluate for

Since the absolute value of any number represents its magnitude from and is therefore always positive, the final answer would be

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Question

If is the greatest integer function, what is the value of ?

Answer

The greatest integer function takes an input and produces the greatest integer less than the input. Thus, the output is always smaller than the input and is an integer itself. Since our input was , we are looking for an integer less than this, which must be since any smaller integer would by definition not be "greatest".

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Question

Let

$y=\left{\begin{matrix} 2x&x\leq 0\ x+9& x>0 \end{matrix}\right.$

What does equal when ?

Answer

Because 3>0 we plug the x value into the bottom equation.

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Question

Let

$y=\left{\begin{matrix} x-9&x\leq 0\ x+9& x>0 \end{matrix}\right.$

What does equal when ?

Answer

Because $0\leq0$ we use the first equation.

Therefore, plugging in x=0 into the above equation we get the following,

.

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Question

Determine the value of if the function is

$f(x)=\left{\begin{matrix} x& x<3\ \pi&x=3 \ 4&x>3 \end{matrix}\right.$

Answer

In order to determine the value of of the function we set

The value comes from the function in the first row of the piecewise function, and as such

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Question

Determine the value of if the function is

$f(x)=\left{\begin{matrix} \sqrt{x}& 0\leq x<4\ \3&x=4 \ e^{-x}&x>4 \end{matrix}\right.$

Answer

In order to determine the value of of the function we set

The value comes from the function in the first row of the piecewise function, and as such

$f(1)=\sqrt{1}=1$

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Question

For the function defined below, what is the value of when ?

$f(x)=\left{\begin{matrix} -x+1&\text{ for }x<1\ x^{2}-2&\text{ for }1< x< 3\ 2^{x}&\text{ for }x\geq 3 \end{matrix}$

Answer

Evaluate the function for . Based on the domains of the three given expressions, you would use $2^{x}$ , since is greater than or equal to .

$2^{x}=2^{3}=8$

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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

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

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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