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Algebra II : Inverse Functions

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Example Questions

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

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

Report an Error

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

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

Report an Error

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

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

Report an Error

Example Question #271 : Introduction To Functions

Relation

Which is true of the relation graphed above?

Possible Answers:

The relation is not a function.

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

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

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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Example Question #801 : Algebra Ii

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

Replace f(x) with :

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

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

Switch the positions of and :

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

or,

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

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

Take the reciprocals of both sides:

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

Multiply both sides by 5:

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

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

Add 7:

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

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

The right expression can be simplified as follows:

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

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

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

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

Report an Error

Example Question #271 : Introduction To Functions

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

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

Possible Answers:

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

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

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

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

$f^{-1}(x)= \ln x e^{3}$

Correct answer:

$f^{-1}(x)= \ln x e^{3}$

Explanation:

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

Replace f(x) with :

$f(x) = e^{x-3}$

$y= e^{x-3}$

Switch the positions of and :

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

$e^{y-3} = x$

Take the natural logarithm of both sides:

$\ln e^{y-3} = \ln x$

By definition, $\ln e^{a}= a$ , so

$y-3 = \ln x$

Add 3 to both sides:

$y-3 +3 = \ln x + 3$

$y = 3 + \ln x$

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

$f^{-1}(x)= 3 + \ln x$

This is not given among the choices; however, remember that by one of the properties of logarithms,

$a = \ln e^{a}$ ,

$f^{-1}(x)= \ln e^{3} + \ln x$

By another property, $\ln M + \ln N = \ln MN$ , so

$f^{-1}(x)= \ln (e^{3} \cdot x)$

$f^{-1}(x)= \ln x e^{3}$ ,

which is among the choices and is the correct answer.

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Algebra II : Inverse Functions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

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 #271 : Introduction To Functions

Example Question #792 : Algebra Ii

Example Question #801 : Algebra Ii

Example Question #271 : Introduction To Functions

All Algebra II Resources

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