Asked by Imani-Jean Johnson on May 11, 2024

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Find the inverse function of f(x) =x+5,x≥−5f ( x ) = \sqrt { x + 5 } , x \geq - 5f(x) =x+5,x5 .

A) f−1(x) =x2−5,x≥0f ^ { - 1 } ( x ) = x ^ { 2 } - 5 , x \geq 0f1(x) =x25,x0
B) f−1(x) =x2−25,x≥0f ^ { - 1 } ( x ) = x ^ { 2 } - 25 , x \geq 0f1(x) =x225,x0
C) f−1(x) =(x−25) 2f ^ { - 1 } ( x ) = ( x - 25 ) ^ { 2 }f1(x) =(x25) 2
D) f−1(x) =(x−5) 2f ^ { - 1 } ( x ) = ( x - 5 ) ^ { 2 }f1(x) =(x5) 2
E) The function does not have an inverse.

Inverse Function

A function that reverses the operations of a given function, such that applying both functions in succession results in the original value.

Square Root

The value that, when multiplied by itself, yields the original number; it is denoted as \(\sqrt{x}\) for a number \(x\).

  • Find the inverse of given functions when possible.
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MC
Meghen ClemensMay 17, 2024
Final Answer :
A
Explanation :
To find the inverse function, swap xxx and yyy and solve for yyy . Starting with y=x+5y = \sqrt{x + 5}y=x+5 , we swap to get x=y+5x = \sqrt{y + 5}x=y+5 . Squaring both sides gives x2=y+5x^2 = y + 5x2=y+5 , and solving for yyy gives y=x2−5y = x^2 - 5y=x25 . The domain of the original function restricts x≥−5x \geq -5x5 , which translates to the range x≥0x \geq 0x0 for the inverse function, making f−1(x)=x2−5,x≥0f^{-1}(x) = x^2 - 5, x \geq 0f1(x)=x25,x0 correct.