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Function  is defined as follows:

\(g(x) = \left\{ \begin{array}{cl} 5x^2 & \text{if } x \le -3, \\ 21+x& \text{if }-3 < x \le 10, \\ 2-\sqrt{x}& \text{if }x > 10. \end{array}\right.\)

 

This function has an inverse.

 What is g^-1(7)

 Apr 30, 2023
 #1
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+1

Since g(x) is a piecewise function, we need to find the inverse of each piece separately.

The inverse of g(x)=5x2 is x=5g(x)​​.

The inverse of g(x)=21+x is x=g(x)−21.

The inverse of g(x)=2−x​ is x=4−4x​.

Therefore, g−1(7) is equal to \begin{align*} \sqrt{\frac{7}{5}} &= \sqrt{\frac{7}{5}} \ 21 + \sqrt{\frac{7}{5}} &= 21 + \sqrt{\frac{7}{5}} \ 4 - 4\sqrt{\frac{7}{5}} &= 4 - 4\sqrt{\frac{7}{5}}. \end{align*}The only solution in the domain of g^{−1}(7) is 4 - 4*sqrt(7/5).

 May 1, 2023
 #2
avatar+118617 
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g-1(7)   does not exist.

 

 May 1, 2023

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