Question #27dca

2 Answers
Mar 22, 2016

See below

Explanation:

a. Given #f(x)=x#
To find #(f(x+h)-f(x))/h# ......(1)

Clearly #f(x+h)=x+h#
Inserting in equation (1), we obtain
#((cancel x+h)-(cancel x))/h#
#h/h=1#, for all values of #h# except #h=0 or oo# where division is not defined as well as #oo/oo# is indeterminate.
b. #lim (f(x+h)-f(x))/h-=f'(x)#
#color(white){W}h->0#

In a. we have already shown that LHS#=1# so long as #h!=0#
Therefore #lim (f(x+h)-f(x))/h=f'(x)=1#
#color(white){WWWW}h->0#
c. From b. #f'(x)=1#
Now #f'(2)# means first derivative of function #f(x)# at #x=2#.
Since first derivative has been found to be constant#=1#, therefore it is independent of the value of #x#.
Hence, #f'(2)=1#.

Mar 22, 2016

To think of it as a power function, recall that #x = x^1#

Explanation:

Applying the power rule, we get

#f'(x) = 1x^(1-1)#

# = 1x^0#

But #x^0 = 1#, so we finish with

#f'(x) = 1#