How do you find the instantaneous rate of change of the function #f(x)=x^2+2x# when x=0?

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Answer 1 · 2018-03-03T06:13:31

#2#.

To find the instantaneous rate of change of a function #f(x)# at a point #d#, we solve for #f'(x)#, and then solve for #f'(d)#.

Here, #f(x)=x^2+2x#

And so #f'(x)# is:

According to the power rule, #d/dxx^n=nx^(n-1)#. We can now, using this, say:

#f'(x)=2x+2#. Here, #d=0#, so we input:

#f'(0)=2*0+2#

#f'(0)=2#

The rate of change of the function at #x=0# is #2#.