Introduction to Functions - High School Math

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Question

is a sine curve. What are the domain and range of this function?

Answer

The domain includes the values that go into a function (the x-values) and the range are the values that come out (the or y-values). A sine curve represent a wave the repeats at a regular frequency. Based upon this graph, the maximum is equal to 1, while the minimum is equal to –1. The x-values span all real numbers, as there is no limit to the input fo a sine function. The domain of the function is all real numbers and the range is $-1\leq f(x)\leq1$ .

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Question

If , which of these values of is NOT in the domain of this equation?

$y = x^2 + \frac{7}{x}$

Answer

Using as the input () value for this equation generates an output () value that contradicts the stated condition of .

Therefore is not a valid value for and not in the equation's domain:

$y = (-1)^2 + \frac{7}{(-1)} \rightarrow y = 1 - 7 \rightarrow y= -6$

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Question

What is the domain of the function?

$f(x)=\sqrt{64-x^2}$

Answer

The domain of a function is the set of possible values for the variable. The range would be the possible values for the solution, .

The value inside of a square root must be greater than or equal to zero in order to have a real solution.

$f(x)=sqrt{64-x^2} ightarrow 64-x^2geq 0$

Now we can solve for in the inequality.

$64geq{x^2}$

$sqrt{64}geqsqrt{x^2}$

Square roots have positive and negative roots, so we need to set up two results. Remember to switch the direction fo the inequality for the negative solution.

These solutions can be combined to give our final answer.

Any values of that are not included will result in an imaginary (impossible) answer.

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Question

Which of the following does NOT belong to the domain of the function $f(x)=\frac{x^{-1}+\sqrt{1-x}}{4x^2+1}$ ?

Answer

The domain of a function includes all of the values of x for which f(x) is real and defined. In other words, if we put a value of x into the function, and we get a result that isn't real or is undefined, then that value won't be in the domain.

If we let x = 0, then we will be forced to evaluate $0^{-1}$ , which is equal to 1/0. The value of 1/0 is not defined, because we can never have zero in a denominator. Thus , because f(0) isn't defined, 0 cannot be in the domain of f(x).

The answer is 0.

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Question

What is the domain of the function:

$f(x)=\frac{1}{x+2}$

Answer

The domain of a function consists of all of the possible values for x. In this case, we want to make sure that we are not dividing by ( in the denominator), since that would make our function undefined. Having would make the denominator . Thus, that is not in our domain. There is nothing else that would make this function undefined, and thus the domain is all real numbers except

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Question

Find the domain of the following function.

$f(x) = \sqrt{x-4}$

Answer

Domain represents all possible values for . In this situation, we want to make sure that the numbers under the square root are greater than or equal to 0. This will be the case when $x - 4 \geq 0$ . Thus, the domain includes all real numbers such that $x \geq 4$ .

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Question

What is the range of $f(x) = x^{2}$ ?

Answer

The range of a function is defined as the possible values for , or the possible outcomes. In this function, it is not possible to get any sort of negative number as an outcome. It is possible to get zero when . Thus, the range is all real numbers greater than or equal to 0.

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Question

What is the range of ?

Answer

The range of a function is all of the possible values that the equation can take. For this equation our value cannot be negative, as a negative number squared still gives us a positive value.

Since , we know that the lowest possible value that can reach is . Therefore the range is $y\geq0$ .

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Question

Which analysis can be performed to determine if an equation is a function?

Answer

The vertical line test can be used to determine if an equation is a function. In order to be a function, there must only be one $\small y$ (or $\small f(x)$ ) value for each value of $\small x$ . The vertical line test determines how many $\small y$ (or $\small f(x)$ ) values are present for each value of $\small x$ . If a single vertical line passes through the graph of an equation more than once, it is not a function. If it passes through exactly once or not at all, then the equation is a function.

The horizontal line test can be used to determine if a function is one-to-one, that is, if only one $\small x$ value exists for each $\small y$ (or $\small f(x)$ ) value. Calculating zeroes, domain, and range can be useful for graphing an equation, but they do not tell if it is a function.

Example of a function:

Example of an equation that is not a function:

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Question

Evaluate $\small f(g(3))$ if and .

Answer

$\small f(g(3))$

This expression is the same as saying "take the answer of $\small g(3)$ and plug it into $\small f(x)$ ."

First, we need to find $\small g(3)$ . We do this by plugging $\small 3$ in for $\small x$ in $\small g(x)$ .

Now we take this answer and plug it into $\small f(x)$ .

We can find the value of $\small f(9)$ by replacing $\small x$ with $\small 9$ .

This is our final answer.

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Question

Let and . What is ?

Answer

THe notation is a composite function, which means we put the inside function g(x) into the outside function f(x). Essentially, we look at the original expression for f(x) and replace each x with the value of g(x).

The original expression for f(x) is . We will take each x and substitute in the value of g(x), which is 2x-1.

We will now distribute the -2 to the 2x - 1.

We must FOIL the term, because .

Now we collect like terms. Combine the terms with just an x.

Combine constants.

The answer is .

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Question

If $f(x)=3x^{2}+5$ and , what is ?

Answer

means gets plugged into .

Thus $g(f(x))= (3x^{2}+5) - 2 = 3x^{2}+3$ .

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Question

Let $f(x)= x^{3}-2x^{2}+x$ and $g(x)=x^{2}-1$ . What is ?

Answer

Calculate and plug it into .

$g(2)= (2)^{2}-1=4-1=3$

$f(3)=(3)^{3}-2(3)^{2}+(3)=27-2(9)+(3)=27-18+3=12$

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Question

Let . What is $f^{-1}(x)$ ?

Answer

We are asked to find $f^{^{-1}} (x)$ , which is the inverse of a function.

In order to find the inverse, the first thing we want to do is replace f(x) with y. (This usually makes it easier to separate x from its function.).

Next, we will swap x and y.

Then, we will solve for y. The expression that we determine will be equal to $f^{^{-1}} (x)$ .

Subtract 5 from both sides.

Multiply both sides by -1.

We need to raise both sides of the equation to the 1/3 power in order to remove the exponent on the right side.

$(5-x)^{\frac{1}{3}}=\left ((1+y)^{3} \right )^{1/3}$

We will apply the general property of exponents which states that $(a^b)^{c}=a^{b\cdot c}$ .

$(5-x)^{\frac{1}{3}}=\left ((1+y)^{3} \right )^{1/3}=(1+y)^{3\cdot \frac{1}{3}}=(1+y)^1=1+y$

Laslty, we will subtract one from both sides.

$(5-x)^{\frac{1}{3}}-1=y$

The expression equal to y is equal to the inverse of the original function f(x). Thus, we can replace y with $f^{-1}(x)$ .

$f^{-1}(x)=(5-x)^{\frac{1}{3}}-1$

The answer is $f^{-1}(x)=(5-x)^{\frac{1}{3}}-1$ .

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Question

What is the inverse of $y=4x^{2}$ ?

Answer

The inverse of requires us to interchange and and then solve for .

$y=4x^{2}$

$x=4y^{2}$

Then solve for :

$y=\frac{\pm \sqrt{x}}{2}$

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Question

If $f(x)=x^{2}+1$ , what is $f^{-1}(x)$ ?

Answer

To find the inverse of a function, exchange the and variables and then solve for .

$x=y^{2}+1$

$x-1=y^{2}$

$\sqrt{x-1}=y=f^{-1}(x)$

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Question

If the function is depicted here, which answer choice graphs ?

Answer

The function shifts a function f(x) units to the left. Conversely, shifts a function f(x) units to the right. In this question, we are translating the graph two units to the left.

To translate along the y-axis, we use the function or .

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