Question

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

Answer 1

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`

Answer 2

`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`