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Given #H = x^2 + 8#, what is #H^-1#?

Answer 1

The notation #H^-1# means the inverse function. The equation of the inverse function can be found by switching the #x# and #y# values inside the function.

#y = x^2 + 8#

#x = y^2 + 8#

#x - 8 = y^2#

#y = +-sqrt(x - 8)#

Hence, #H^-1(x) = +-sqrt(x- 8)#.

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Answer 2

Madelyn Anderson

To find the inverse of a function, you switch the roles of the variables ( x ) and ( y ) and then solve for ( y ). So, given ( H = x^2 + 8 ), to find ( H^{-1} ), we switch ( x ) and ( y ) to get ( x = y^2 + 8 ). Then we solve for ( y ).

Starting with ( x = y^2 + 8 ), we isolate ( y ): [ x - 8 = y^2 ] [ \sqrt{x - 8} = y ]

So, ( H^{-1} = \sqrt{x - 8} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.