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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