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

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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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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.

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.

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.

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