What is the instantaneous rate of change of $f(x)=sqrt(3x^2-4x+5)$ at $x=-1$ ?

Question

2054 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-07T00:04:06

It is $f'(-1)=-5/[2sqrt3]$

The rate of change is the value of the first derivative of the function $f(x)$ at $x=-1$

Hence

$f'(x)=[3x-2]/sqrt[3x^2-4x+5]$

so

$f'(-1)=-5/[2sqrt3]$

What is the instantaneous rate of change of f(x)=sqrt(3x^2-4x+5) f(x)=sqrt(3x^2-4x+5) at x=-1 x=-1 ?