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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #69 : Inverse Functions

Determine the inverse: x+2y = -4

Possible Answers:

$y=-\frac{1}{2}x -2$

$y=-2x -\frac{1}{2}$

y=-2x -4

$y=-2x -\frac{1}{4}$

$y=-\frac{1}{2}x -4$

Correct answer:

y=-2x -4

Explanation:

In order to solve for the inverse, interchange the x and the y variables, and solve for y.

y+2x = -4

Subtract from both sides of the equation.

y+2x-2x =-2x -4

The answer is: y=-2x -4

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Example Question #70 : Inverse Functions

Relation

Above is the graph of a function f(x) . Which choice gives the graph of $f^{-1}(x)$ ?

Possible Answers:

Relation

Correct answer:

Relation

Explanation:

Given the graph of f(x) , the graph of its inverse, $f^{-1}(x)$ is the reflection of the former about the line y = x . This line is in dark green below; critical points are reflected as shown:

Relation

The blue figure is $f^{-1}(x)$ , recreated below:

Relation

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Example Question #71 : Inverse Functions

Define a function $f(x) = \frac{1}{5}x +7$ .

Which statement correctly gives $f^{-1}(x)$ ?

Possible Answers:

$f^{-1}(x) = \frac{5}{7}x-7$

$f^{-1}(x) = \frac{5}{7}x-35$

None of the other choices gives the correct response.

$f^{-1}(x) =5x-7$

$f^{-1}(x) =5x-35$

Correct answer:

$f^{-1}(x) =5x-35$

Explanation:

The inverse function $f^{-1}(x)$ of a function f(x) can be found as follows:

Replace f(x) with :

$f(x) = \frac{1}{5}x +7$

$y= \frac{1}{5}x +7$

Switch the positions of and :

$x= \frac{1}{5}y +7$

or,

$\frac{1}{5}y +7 = x$

Solve for - that is, isolate it on one side.

Subtract 7:

$\frac{1}{5}y +7 - 7 = x - 7$

$\frac{1}{5}y = x - 7$

Multiply by 5, distributing on the right:

$5 \cdot \frac{1}{5}y =5 \cdot (x - 7)$

$y =5 \cdot x - 5 \cdot 7$

y =5x-35

Replace with $f^{-1}(x)$ :

$f^{-1}(x) =5x-35$

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Example Question #72 : Inverse Functions

Define a function $f(x) = 5^{x} + 5$ .

Which statement correctly gives $f^{-1}(x)$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The inverse function $f^{-1}(x)$ of a function f(x) can be found as follows:

Replace f(x) with :

$f(x) = 5^{x} + 5$

$y = 5^{x} + 5$

Switch the positions of and :

$x = 5^{y} + 5$ ,

$5^{y} + 5 = x$

Solve for - that is, isolate it on one side - as follows:

First, subtract 5 from both sides:

$5^{y} + 5- 5 = x - 5$

$5^{y} = x - 5$

Take the base-5 logarithm of both sides:

$\log_{5} 5^{y} = \log_{5} (x - 5)$

A property of logarithms states that $\log_{a} a^{N}= N$ , so

$y = \log_{5} (x - 5)$

Replace with $f^{-1}(x)$ :

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

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Example Question #71 : Inverse Functions

Define a function $f(x) = \frac{x-2}{2x}$ .

Which statement correctly gives $f^{-1}(x)$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The inverse function $f^{-1}(x)$ of a function f(x) can be found as follows:

Replace f(x) with :

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

$y= \frac{x-2}{2x}$

Switch the positions of and :

$x= \frac{y-2}{2y}$

Solve for - that is, isolate it on one side - as follows:

Split the expression at right into the difference of two separate expressions:

$x = \frac{y}{2y} - \frac{2}{2y}$

Simplify:

$x = \frac{1}{2} - \frac{1}{y}$

Add $\frac{1}{y} - x$ to both sides:

$x + \frac{1}{y} - x = \frac{1}{2} - \frac{1}{y} + \frac{1}{y} - x$

$\frac{1}{y} = \frac{1}{2} - x$

Simplify the expression at right:

$\frac{1}{y} = \frac{1}{2} - \frac{2x}{2}$

$\frac{1}{y} = \frac{1-2x}{2}$

Take the reciprocal of both sides:

$y= \frac{2}{1-2x}$

Replace with $f^{-1}(x)$ :

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

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Example Question #272 : Introduction To Functions

Define a function $f(x) = \sqrt[3]{4x-9}$ .

Which statement correctly gives $f^{-1}(x)$ ?

Possible Answers:

$f^{-1}(x) =\frac{1}{4} x^{3} + \frac{729}{4}$

$f^{-1}(x) =\frac{1}{4} x^{3} + \frac{27}{4}x^{2}+\frac{243}{4}x+ \frac{729}{4}$

$f^{-1}(x) =\frac{1}{4} x^{3} + \frac{9}{4}$

$f^{-1}(x) =\frac{1}{4} x^{3} - \frac{27}{4}x^{2}+\frac{243}{4}x- \frac{729}{4}$

$f^{-1}(x) =\frac{1}{4} x^{3} + 9$

Correct answer:

$f^{-1}(x) =\frac{1}{4} x^{3} + \frac{9}{4}$

Explanation:

The inverse function $f^{-1}(x)$ of a function f(x) can be found as follows:

Replace f(x) with :

$f(x) = \sqrt[3]{4x-9}$

$y = \sqrt[3]{4x-9}$

Switch the positions of and :

$x = \sqrt[3]{4y-9}$

$\sqrt[3]{4y-9} = x$

Solve for - that is, isolate it on one side - as follows:

Raise both sides to the third power:

$\left (\sqrt[3]{4y-9} \right )^{3}= x^{3}$

$4y-9 = x^{3}$

Add 9 to both sides:

$4y-9 + 9 = x^{3} + 9$

$4y = x^{3} + 9$

Multiply both sides by $\frac{1}{4}$ , distributing on the right side:

$\frac{1}{4} \cdot 4y =\frac{1}{4} \cdot ( x^{3} + 9)$

$y =\frac{1}{4} \cdot x^{3} + \frac{1}{4} \cdot 9$

$y =\frac{1}{4} x^{3} + \frac{9}{4}$

Replace with $f^{-1}(x)$ :

$f^{-1}(x) =\frac{1}{4} x^{3} + \frac{9}{4}$

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Example Question #71 : Inverse Functions

Define a function $f(x)= \frac{x}{x-1}$ .

True or false: f(x) is its own inverse.

Possible Answers:

False

True

Correct answer:

True

Explanation:

The inverse function $f^{-1}(x)$ of a function f(x) can be found as follows:

Replace f(x) with :

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

$y = \frac{x}{x-1}$

Switch and :

$x = \frac{y}{y-1}$

Solve for - that is, isolate on one side of the equation - as follows:

Multiply both sides by y -1 , distributing on the right side:

$x\cdot (y-1) = \frac{y}{y-1} \cdot (y-1)$

$x\cdot y-x\cdot 1 = y$

x y-x = y

Add x-y to both sides to get all terms to the left, then factor out :

x y-x+ x- y = y + x- y

x y - y =x

$x \cdot y -1 \cdot y =x$

(x -1 ) y =x

Divide both sides by x-1 :

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

$y =\frac{x}{x -1 }$

Replace with $f^{-1}(x)$ :

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

Therefore, $f^{-1}(x)= f(x)$ , and f(x) is indeed its own inverse.

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Example Question #274 : Introduction To Functions

Define a function $f(x) = 3 e ^{x} + 4$ .

Which statement correctly gives $f^{-1}(x)$ ?

Possible Answers:

$f^{-1}(x)=\ln \left (\frac{1}{3} x - \frac{4}{3} \right )$

$f^{-1}(x)=\ln \left (\frac{1}{3} x -4 \right )$

$f^{-1}(x)=\log \left (\frac{1}{3} x - \frac{4}{3} \right )$

$f^{-1}(x)=\log \left (\frac{1}{3} x -4 \right )$

None of the other choices gives the correct response.

Correct answer:

$f^{-1}(x)=\ln \left (\frac{1}{3} x - \frac{4}{3} \right )$

Explanation:

The inverse function $f^{-1}(x)$ of a function f(x) can be found as follows:

Replace f(x) with :

$f(x) = 3 e ^{x} + 4$

$y= 3 e ^{x} + 4$

Switch the positions of and :

$x= 3 e ^{y} + 4$

$3 e ^{y} + 4 = x$

Solve for - that is, isolate it on one side.

First, subtract 4:

$3 e ^{y} + 4 - 4 = x - 4$

$3 e ^{y} = x - 4$

Multiply by $\frac{1}{3}$ and distribute on the right:

$\frac{1}{3} \cdot 3 e ^{y} =\frac{1}{3} \cdot (x - 4)$

$e ^{y} =\frac{1}{3} \cdot x - \frac{1}{3} \cdot 4$

$e ^{y} =\frac{1}{3} x - \frac{4}{3}$

Take the natural logarithm of both sides:

$\ln e ^{y} =\ln \left (\frac{1}{3} x - \frac{4}{3} \right )$

$y =\ln \left (\frac{1}{3} x - \frac{4}{3} \right )$

Replace with $f^{-1}(x)$ :

$f^{-1}(x)=\ln \left (\frac{1}{3} x - \frac{4}{3} \right )$

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Example Question #71 : Inverse Functions

Relation

Which is true of the relation graphed above?

Possible Answers:

The relation is a function, but it does not have an inverse.

The relation is a function, and it has an inverse.

The relation is not a function.

Correct answer:

The relation is not a function.

Explanation:

A relation is a function if and only if it passes the Vertical Line Test (VLT) - that is, no vertical line exists that passes through its graph more than once. From the diagram below, we see that at least one such line exists:

Relation

The relation fails the VLT, so it is not a function.

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Example Question #792 : Algebra Ii

Untitled

The above table shows a function with domain $\left \{ -3, -2, -1, 0, 1, 2, 3 \right \}$ .

True or false: f(x) has an inverse function.

Possible Answers:

True

False

Correct answer:

True

Explanation:

A function f(x) has an inverse function if and only if, for all a,b in the domain of f(x) , if $a \ne b$ , it follows that $f(a) \ne f(b)$ . In other words, no two values in the domain can be matched with the same range value.

If we order the rows by range value, we see this to be the case:

Untitled

If follows that f(x) has an inverse function.

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #69 : Inverse Functions

Example Question #70 : Inverse Functions

Example Question #71 : Inverse Functions

Example Question #72 : Inverse Functions

Example Question #71 : Inverse Functions

Example Question #272 : Introduction To Functions

Example Question #71 : Inverse Functions

Example Question #274 : Introduction To Functions

Example Question #71 : Inverse Functions

Example Question #792 : Algebra Ii

All Algebra II Resources

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