Let $g ( t ) = \frac { 4 } { t } + 6$

Let $g ( t ) = \frac { 4 } { t } + 6$ . Find g ^{- 1} (t) .

A) $g ^ { - 1 } ( t ) = \frac { 4 } { t - 6 }$
B) $g ^ { - 1 } ( t ) = \frac { 4 } { t } + 6$
C) $g ^ { - 1 } ( t ) = \frac { t } { 4 } - 6$
D) $g ^ { - 1 } ( t ) = \frac { 7 } { t } - 6$
E) $g ^ { - 1 } ( t ) = \frac { 6 } { t - 4 }$

Correct Answer:

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