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3 years ago

What is the inverse of f(x)= cube root of x +2.

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

3 0

$f(x) = \sqrt[3]{x + 2} \\y = \sqrt[3]{x + 2} \\y^{3} = x + 2 \\x^{3} = y + 2 \\x^{3} - 2 = y \\x^{3} - 2 = f^{-1}(x)$

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Describe the graph of y={1/(2x-10)}-3 compared to the graph of y=1/x

tester [92]

$\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)= A( Bx+ C)+ D
\\\\
~~~~y= A( Bx+ C)+ D
\\\\
f(x)= A\sqrt{ Bx+ C}+ D
\\\\
f(x)= A(\mathbb{R})^{ Bx+ C}+ D
\\\\
f(x)= A sin\left( B x+ C \right)+ D
\\\\
--------------------$

$\bf \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\
\bullet \textit{ flips it upside-down if } A\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if } B\textit{ is negative}$

$\bf ~~~~~~\textit{reflection over the y-axis}
\\\\
\bullet \textit{ horizontal shift by }\frac{ C}{ B}\\
~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}\\\\
~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by } D\\
~~~~~~if\ D\textit{ is negative, downwards}\\\\
~~~~~~if\ D\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{ B}$

with that template in mind, let's check these two

$\bf \stackrel{parent}{y=\cfrac{1}{x}}\qquad \qquad\qquad \qquad \stackrel{transformed}{y=\cfrac{1}{\stackrel{B}{2}x\stackrel{C}{-10}}\stackrel{D}{-3}}\\\\
-------------------------------\\\\
B=2\qquad \textit{shrinks horizontally by }\frac{1}{2}
\\\\\\
C=-10\qquad \cfrac{C}{B}=\cfrac{-10}{2}\implies -5\qquad \textit{horizontally right-shifted by }5
\\\\\\
D=-3\qquad \textit{vertically down-shifted by }3$

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