What is the instantaneous rate of change of f(x)=x/(-x-8)f(x)=x−x−8 at x=4 x=4?
2 Answers
Explanation:
"the instantaneous rate of change at x = 4"the instantaneous rate of change at x = 4
"is the value of the derivative at x = 4"is the value of the derivative at x = 4
"differentiate using the "color(blue)"quotient rule"differentiate using the quotient rule
"given "f(x)=(g(x))/(h(x))" then"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
Explanation:
"Original Function:" \qquad f(x) = x / {-x-8} "Rewrite a little:" \qquad f(x) = - x / {x+8} "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 ]. "Summary:" \qquad \qquad \qquad \quad \ f'(x) = - { 8 }/ [ (x+8)^2 ]. "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. "Summarize:" \qquad \qquad \qquad \quad \quad \ f'(4) = - 1/18.