Use the function f ( x ) = 1 x f ( x ) = \frac { 1 } { x } f ( x ) = x 1 to find and simplify the expression for f ( 14 + h ) − f ( 14 ) h \frac { f ( 14 + h ) - f ( 14 ) } { h } h f ( 14 + h ) − f ( 14 ) .A) − 1 14 ( h + 14 ) , h ≠ 0 - \frac { 1 } { 14 ( h + 14 ) } , h
eq 0 − 14 ( h + 14 ) 1 , h  = 0 B) − 14 h ( h + 14 ) , h ≠ 0 - \frac { 14 h } { ( h + 14 ) } , h
eq 0 − ( h + 14 ) 14 h , h  = 0 C) − h 14 ( h + 14 ) , h ≠ 0 - \frac { h } { 14 ( h + 14 ) } , h
eq 0 − 14 ( h + 14 ) h , h  = 0 D) 1 14 ( h + 14 ) , h ≠ 0 \frac { 1 } { 14 ( h + 14 ) } , h
eq 0 14 ( h + 14 ) 1 , h  = 0 E) − 1 ( h + 14 ) , h ≠ 0 - \frac { 1 } { ( h + 14 ) } , h
eq 0 − ( h + 14 ) 1 , h  = 0

Question

Use the function   f ( x )  = 1 x f ( x )  = \frac { 1 } { x } f ( x )  = x 1 ​   to find and simplify the expression for   f ( 14 + h )  − f ( 14 )  h \frac { f ( 14 + h )  - f ( 14 )  } { h } h f ( 14 + h )  − f ( 14 )  ​   .A)    − 1 14 ( h + 14 )  , h ≠ 0 - \frac { 1 } { 14 ( h + 14 )  } , h 
eq 0 − 14 ( h + 14 )  1 ​ , h  = 0 B)    − 14 h ( h + 14 )  , h ≠ 0 - \frac { 14 h } { ( h + 14 )  } , h 
eq 0 − ( h + 14 )  14 h ​ , h  = 0 C)    − h 14 ( h + 14 )  , h ≠ 0 - \frac { h } { 14 ( h + 14 )  } , h 
eq 0 − 14 ( h + 14 )  h ​ , h  = 0 D)    1 14 ( h + 14 )  , h ≠ 0 \frac { 1 } { 14 ( h + 14 )  } , h 
eq 0 14 ( h + 14 )  1 ​ , h  = 0 E)    − 1 ( h + 14 )  , h ≠ 0 - \frac { 1 } { ( h + 14 )  } , h 
eq 0 − ( h + 14 )  1 ​ , h  = 0

Hallie Moffitt · Accepted Answer

Final Answer : A, Explanation :   Substituting   f ( 14 + h ) = 1 14 + h f(14+h) = \frac{1}{14+h} f ( 14 + h ) = 14 + h 1 ​   and   f ( 14 ) = 1 14 f(14) = \frac{1}{14} f ( 14 ) = 14 1 ​   into the expression and simplifying gives   − 1 14 ( h + 14 ) -\frac{1}{14(h+14)} − 14 ( h + 14 ) 1 ​  , with   h ≠ 0 h 
eq 0 h  = 0   to avoid division by zero.

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