Algebraic Concepts - HiSet: High School Equivalency Test: Math

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Question

What is the slope of the line given by the following table?

$\begin{tabular}{ | c | c |} \hline x & y \ \hline 0 & 2 \ \hline 1 & 3 \ \hline 2 & 4 \ \hline \end{tabular}$

Answer

Given two points

$(x_{0},y_{0})$ and $(x_{1},y_{1})$

the formula for a slope is

$m=\frac{y_{1}-y_{0}}{x_{1}-x_{0}}$ .

Thus, since our given table is

$\begin{tabular}{ | c | c |} \hline x & y \ \hline 0 & 2 \ \hline 1 & 3 \ \hline 2 & 4 \ \hline \end{tabular}$

we select two points, say

and

and use the slope formula to compute the slop.

Thus,

$m = \frac{3-2}{1-0}=\frac{1}{1} = 1$ .

Hence, the slope of the line generated by the table is

.

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Question

Define $f(x) = x^{2}+ x$ .

Give the average rate of change of over the interval .

Answer

The average rate of change of a function over an interval is equal to

$\frac{f(b)-f(a)}{b-a}$

Setting , this is

$\frac{f(3)-f(2)}{3-2} = \frac{f(3)-f(2)}{1} = f(3)-f(2)$

Evaluate and by substitution:

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

$f(3) = 3^{2}+ 3= 9+ 3 = 12$

$f(2) = 2^{2}+ 2 = 4+ 2 = 6$

$\frac{f(3)-f(2)}{3-2} = 12 - 6 = 6$ ,

the correct response.

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Question

Define $f(x) = \left{\begin{matrix} x^{2}+x ,&x \le 0 \ x^{2}-x, & x> 0\end{matrix}\right.$ .

Give the average rate of change of over the interval .

Answer

The average rate of change of a function over an interval is equal to

$\frac{f(b)-f(a)}{b-a}$

Setting , this is

$\frac{f(1)-f(-1)}{1-(-1)} = \frac{f(1)-f(-1)}{2}$

Evaluate using the definition of for :

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

$f(1) = 1^{2}-1 = 1 - 1 = 0$

Evaluate using the definition of for $x\le 0$ :

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

$f(-1) = (-)1^{2}+ (-1) = 1 - 1 = 0$

The average rate of change is therefore

$\frac{f(1)-f(-1)}{2} = \frac{0-0}{2} = 0$ .

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Question

Identify the coefficients in the following formula:

$\frac{5x^2+2x-4y^3+3y-6}{7\times12+9-13}=29$

Answer

Generally speaking, in an equation a coefficient is a constant by which a variable is multiplied. For example, and are coefficients in the following equation:

In our equation, the following numbers are coefficients:

$5,\ 2,\ 4,\ \textup{and }3$

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Question

What is the coefficient of the second highest term in the expression: ?

Answer

Step 1: Rearrange the terms from highest power to lowest power.

We will get: .

Step 2: We count the second term from starting from the left since it is the second highest term in the rearranged expression.

Step 3: Isolate the term.

The second term is

Step 4: Find the coefficient. The coefficient of a term is considered as the number before any variables. In this case, the coefficient is .

So, the answer is .

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Question

Define functions $f(x) = \sqrt{x}$ and $g(x) =- x^{2}$

Give the domain of the composition $g\circ f$ .

Answer

The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that falls in the domain of .

$f(x) = \sqrt{x}$ , a square root function, whose domain is the set of all values of that make the radicand nonnegative. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

$g(x) =- x^{2}$ , a polynomial function, whose domain is the set of all real numbers. Therefore, the domain is not restricted further by . The domain of $g\circ f$ is $\left [ 0, \infty\right )$ .

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Question

Define functions $f(x) = \sqrt{x}$ and $g(x) =- x^{2}$

Give the domain of the composition $f \circ g$ .

Answer

The domain of the composition of functions $f \circ g$ is the set of all values that fall in the domain of such that falls in the domain of .

$g(x) =- x^{2}$ , a polynomial function, so the domain of is equal to the set of all real numbers, $\left ( -\infty, \infty\right )$ . Therefore, places no restrictions on the domain of $f \circ g$ .

$f(x) = \sqrt{x}$ , a square root function, whose domain is the set of all values of that make the radicand nonnegative. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ . Therefore, we must select the set of all in $\left ( -\infty, \infty\right )$ such that $g(c) \ge 0$ , or, equivalently,

$-c^{2} \ge 0$

However, $c^{2} \ge 0$ , so $-c^{2} \le 0$ , with equality only if . Therefore, the domain of $f \circ g$ is the single-element set $\left { 0 \right }$ , which is not among the choices.

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Question

Define functions $f(x) = \frac{1}{x-1}$ and $g(x)= \sqrt{x}+ 1$ .

Give the domain of the composition $g\circ f$ .

Answer

The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that falls in the domain of .

$f(x) = \frac{1}{x-1}$ , a rational function, so its domain is the set of all numbers except those that make its denominator zero. if and only if , making its domain $(-\infty, 1) \cup (1, \infty)$ .

$g(x)= \sqrt{x}+ 1$ , a square root function, so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

Therefore, we need to further restrict the domain to those values in $(-\infty, 1) \cup (1, \infty)$ that make

$f(c) \ge 0$

Equivalently,

$\frac{1}{c-1}\ge 0$

Since 1 is positive, and the quotient is nonnegative as well, it follows that

,

and

.

The domain of $g\circ f$ is therefore $(1, \infty)$ .

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Question

Define functions $f(x) = \frac{1}{x-1}$ and $g(x)= \sqrt{x}+ 1$ .

Give the domain of the composition $f \circ g$ .

Answer

The domain of the composition of functions $f \circ g$ is the set of all values that fall in the domain of such that falls in the domain of .

$g(x)= \sqrt{x}+ 1$ , a square root function, so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

$f(x) = \frac{1}{x-1}$ , a rational function, so its domain is the set of all numbers except those that make its denominator zero. if and only if , so from $\left [ 0, \infty\right )$ , we exclude only the value so that

;

that is,

$\sqrt{c}+ 1 = 1$

$\sqrt{c}+ 1 - 1 = 1 - 1$

$\sqrt{c} = 0$

Excluding this value from the domain, this leaves $\left ( 0, \infty\right )$ as the domain of $f \circ g$ .

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Question

Define functions $f(x) = x^{2}-16$ and $g(x) = \sqrt{x}$ .

Give the domain of the composition $g\circ f$ .

Answer

The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that falls in the domain of .

$f(x) = x^{2}-16$ , a polynomial function, so its domain is the set of all real numbers.

$g(x) = \sqrt{x}$ , so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

Therefore, the domain of $g\circ f$ is the set of all real numbers such that

$f(c) \ge 0$ .

Equivalently,

$c^{2}-16 \ge 0$

To solve this, first, solve the corresponding equation:

$c^{2}-16 = 0$

Factor the polynomial at left using the difference of squares pattern:

By the Zero Factor Property, either

, in which case ,

or

, in which case .

These are the boundary values of the intervals to be tested. Choose an interior value of each interval and test it in the inequality:

$(-\infty, -4)$

Testing :

$(-5)^{2}-16 \ge 0$

$25-16 \ge 0$

$9 \ge 0$ - True; include

Testing :

$0^{2}-16 \ge 0$

$0 -16 \ge 0$

$-16 \ge 0$ - False; exclude

$( 4 , \infty)$

Testing :

$5^{2}-16 \ge 0$

$25-16 \ge 0$

$9 \ge 0$ - True; include

Include $(-\infty, -4)$ and $( 4 , \infty)$ . Also include the boundary values, since equality is included in the inequality statement ( $\ge$ ).

The correct domain is $(-\infty, -4] \cup [4 , \infty)$ .

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Question

Define functions $f(x) = x^{2}-16$ and $g(x) = \sqrt{x}$ .

Give the domain of the composition $f \circ g$ .

Answer

The domain of the composition of functions $f \circ g$ is the set of all values that fall in the domain of such that falls in the domain of .

$g(x) = \sqrt{x}$ , a square root function, so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

$f(x) = x^{2}-16$ , a polynomial function; its domain is the set of all reals, so it does not restrict the domain any further.

The correct domain is $\left [ 0, \infty\right )$ .

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Question

Define functions $f(x) = \sqrt{x-8}$ and $g(x) = \sqrt{x+ 8}$ .

Give the domain of the function .

Answer

The domain of the sum of two functions is the intersections of the domains of the individual functions. Both and are square root functions, so their radicands must both be positive.

The domain of is the set of all values of such that:

$x-8 \ge 0$

Adding 8 to both sides:

$x-8 + 8 \ge 0 + 8$

$x \ge 8$

This domain is $[8, \infty )$ .

The domain of is the set of all values of such that:

$x+8 \ge 0$

Subtracting 8 tfrom both sides:

$x+ 8-8 \ge 0 - 8$

$x \ge -8$

This domain is $[-8, \infty )$

The domain of is the intersection of the domains, which is $[8, \infty )$ .

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Question

Define functions $f(x) = \sqrt{x+8}$ and $g(x) = \sqrt{8-x}$ .

Give the domain of the function .

Answer

The domain of the difference of two functions is the intersections of the domains of the individual functions. Both and are square root functions, so their radicands must both be positive.

The domain of is the set of all values of such that:

$x+8 \ge 0$

Subtracting 8 tfrom both sides:

$x+ 8-8 \ge 0 - 8$

$x \ge -8$

This domain is $[-8, \infty )$

The domain of is the set of all values of such that:

$8-x \ge 0$

Adding to both sides:

$8-x + x \ge 0 + x$

$8 \ge x$ , or

$x \le 8$

This domain is $(-\infty , 8]$ .

The domain of is the intersection of these, .

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Question

Define functions $f(x) = \sqrt[3]{x-8}$ and $g(x) = \sqrt[3]{x+ 8}$ .

Give the domain of the function .

Answer

The domain of the sum of two functions is the intersections of the domains of the individual functions.

Both and are cube root functions. There is no restriction on the value of the radicand of a cube root, so both of these functions have as their domain the set of all real numbers $(-\infty, \infty)$ . It follows that also has domain $(-\infty, \infty)$ .

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Question

Define

$f(x)= \left{\begin{matrix} x-2,&x< 0 \ x+ 3,& x \ge 0\end{matrix}\right.$

Give the range of the function.

Answer

The range of a function is the set of all possible values of over its domain.

Since this function is piecewise-defined, it is necessary to examine both parts of the function and extract the range of each.

For , it holds that , so

,

or

$x \in(-\infty, -2)$ .

For $x \ge 0$ , it holds that , so

$x \ge 0$

$x+ 3 \ge 0+ 3$

$f(x) \ge 3$ ,

or

$x \in [3, \infty)$

The overall range of is the union of these sets, or $(-\infty, -2) \cup [3, \infty)$ .

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Question

The graph of a function is given above, with the coordinates of two points on the curve shown. Use the coordinates given to approximate the rate of change of the function between the two points.

Answer

The rate of change between two points on a curve can be approximated by calculating the change between two points.

Let be the coordinates of the first point and be the coordinates of the second point. Then the formula giving approximate rate of change is:

$\frac{y_2 - y_1} {x_2 - x_1}$

Notice that the numerator is the overall change in y, and the denominator is the overall change in x.

The calculation for the problem proceeds as follows:

Let be the first point and be the second point. Substitute in the values from these coordinates:

$\frac{y_2 - y_1} {x_2 - x_1} \rightarrow \frac {5-2}{8-3}$

Subtract to get the final answer:

$\frac{3}{5}$

Note that it does not matter which you assign to be the first point and which you assign to be the second, as it will lead to the same value due to negatives canceling out.

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Question

Above is the graph of a function. The average rate of change of over the interval is $-\frac{4}{5}$ . Which of these values comes closest to being a possible value of ?

Answer

The average rate of change of a function on the interval is equal to

$r = \frac{f(b)-f(a)}{b - a}$ .

Restated, it is the slope of the line that passes through and .

To find the correct value of that answers this question, it suffices to examine the line with slope $-\frac{4}{5}$ through and find the point among those given that is closest to the line. This line falls 4 units for every 5 horizontal units, so the line looks like this:

The -coordinate of the point of intersection is closer to 2 than to any other of the values in the other four choices. This makes 2 the correct choice.

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Question

Above is the graph of a function , which is defined and continuous on $(-\infty, \infty)$ . The average rate of change of on the interval is 4. Estimate .

Answer

The rate of change of a function on the interval is equal to

$r = \frac{f(b)-f(a)}{b - a}$ .

Set . Examine the figure below:

The graph passes through the point , so . Therefore,

$4 = \frac{f(5)-f(1)}{5-1}$

and, substituting,

$4 = \frac{f(5)-2}{4}$

Solve for using algebra:

$\frac{f(5)-2}{4} \cdot 4 = 4 \cdot 4$

,

the correct response.

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Question

Above is the graph of a function . Estimate the rate of change of on the interval

Answer

The rate of change of a function on the interval is equal to

$r = \frac{f(b)-f(a)}{b - a}$ .

Set . Refer to the graph of the function below:

The graph passes through and .

. Thus,

$r = \frac{f(-1.5)-f(-2)}{-1.5 - (-2)} = \frac{5-2}{0.5} = \frac{3}{0.5} = 6$ ,

the correct response.

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Question

Define $f(x)= -\frac{1}{x+1}$ and .

Evaluate $(f\circ g )(2)$ .

Answer

By definition,

$(f\circ g )(2) = f(g(2))$ ,

so first, evaluate by substituting 2 for in :

$(f\circ g )(2) = f(4)$ ,

so evaluate through a similar substitution:

$f(x)= -\frac{1}{x+1}$

$f(4)= -\frac{1}{4+1} = - \frac{1}{5}$ ,

the correct response.

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