What is the instantaneous rate of change of #f(x)=1/(x^2-x+3 )# at #x=0 #?

2 Answers
Timber Lin
Jan 20, 2018

1/9

Explanation:

find #f'(x)# or the instantaneous rate of change of #f(x)# at x.

#=-1/((x^2-x+3)^2)*d/dx(x^2-x+3)#

(power rule: #d/dx(x^n)=nx^(n-1)# and chain rule: #d/dx(f(g(x)))=f'(g(x))g'(x)#)

#=-1/((x^2-x+3)^2)*(2x-1)#

plug in 0 for x:

#f'(0)=-1/((0^2-0+3)^2)*(2(0)-1)#

#f'(0)=-1/(3^2)*(-1)#

#f'(0)=1/9#

Jim G.
Jan 20, 2018

#1/9#

Explanation:

#"the instantaneous rate of change of f(x) at x=0"#
#"is f'(0)"#

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

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

#f'(x)=g'(h(x))xxh'(x)larrcolor(blue)"chain rule"#

#f(x)=1/(x^2-x+3)=(x^2-x+3)^-1#

#rArrf'(x)=-(x^2-x+3)^-2xxd/dx(x^2-x+3)#

#color(white)(rArrf'(x))=-(2x-1)/(x^2-x+3)^2#

#rArrf'(0)=-(-1)/3^2=1/9#

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