How do you find the instantaneous rate of change for #f(x)=(x^2-2)/(x-1)# for x=2?

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Answer 1 · 2018-02-02T14:59:00

Please refer to the explanation below.

We have to take the derivative of the function #f(x)#, and then plug in #x=2# into #f'(x)#.

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

So, we have to use the quotient rule, which states that,

#d/dx(u/v)=(u'v-uv')/v^2#

Here, #u=x^2-2#, #v=x-1#

#:.f'(x)=(2x(x-1)-1*(x^2-2))/(x-1)^2#

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

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

Plugging in #x=2#, we get

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

#f'(2)=(4-4+2)/1^2#

#f'(2)=2/1#

#f'(2)=2#

So, the instantaneous rate of change will be #2#.

Answer 2 · 2018-02-02T15:01:42

#m=2#

By finding the derivativ of #f(x)#

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

Now we determinate the value of the divertive at the point x=2

#f'(2)=2#