Question

Question #27dca

Answer 1

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`.

Answer 2

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`