What is the instantaneous rate of change of #f(x)=x/(-x-8)# at #x=4 #?

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Feb 9, 2018

Answer:

# \ #

# "Answer is:" \qquad \quad - 1/18. #

Explanation:

# \ #

# "The required quantity is the derivative of" \ \ f(x) \ \ "evaluated at" #
# \qquad \qquad x = 4. #

# "So, let's calculate the derivative of" \ \ f(x): #

  1. # "Original Function:" \qquad f(x) = x / {-x-8} #
  2. # "Rewrite a little:" \qquad f(x) = - x / {x+8}#
  3. # "Quotient Rule:" \qquad f'(x) = - { (x+8) [ x ]' - x [x+8]' }/ [ (x+8)^2 ] #
    # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = - { (x+8) [ 1 ] - x [1+0] }/ [ (x+8)^2 ] #
    # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = - { (x+8) - x }/ [ (x+8)^2 ] #
    # \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = - { 8 }/ [ (x+8)^2 ]. #
  4. # "Summary:" \qquad \qquad \qquad \quad \ f'(x) = - { 8 }/ [ (x+8)^2 ]. #
  5. # "Evaluate" \ \ f'(x) \ \ "at" \ \ x = 4, #
    # \qquad \quad \quad "to get the instanteous rate of change at" \ \ x=4: #
    # f'(4) = - { 8 }/ [ (4+8)^2 ] \ = - \ 8/12^2 \ = - \ 8/ ( 12 \cdot 12 ) #
    # \qquad\quad \quad \ = - \color{red}{ cancel {8} }/ ( color{red}{ cancel {2} } \cdot 6 \cdot color{red}{ cancel {4} } \cdot 3 ) = \ \ - 1/18. #
  6. # "Summarize:" \qquad \qquad \qquad \quad \quad \ f'(4) = - 1/18. #

# :. "instantaneous rate of change of"\ \ f(x) \ \ "at" \ (x=4) \ = - 1/18.#

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Jim G. Share
Feb 9, 2018

Answer:

#-1/18#

Explanation:

#"the instantaneous rate of change at x = 4"#

#"is the value of the derivative at x = 4"#

#"differentiate using the "color(blue)"quotient rule"#

#"given "f(x)=(g(x))/(h(x))" then"#

#f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#

#g(x)=xrArrg'(x)=1#

#h(x)=-x-8rArrh'(x)=-1#

#rArrf'(x)=(-x-8-x(-1))/(-x-8)^2=-8/(-x-8)^2#

#rArrf'(4)=-8/(144)=-1/18#

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