Question

Use the difference quotient to estimate the instantaneous rate of change in f(x)=`x^2` at x=3?

Answer 1

See below

Explanation:

`f(x) =x^2`

Let `f'(3) =`the instantaneous rate of change in f(x) at x=3

To estimate `f'(3)` we can use the difference quotient:

`(f(x+deltax)-f(x))/(deltax)` for `x=3` and some small `deltax`

Let's choose `deltax = 0.01`

Then, `f'(3) approx ((3+0.01)^2- 3^2)/0.01`

`approx (9.0601-9)/0.01 approx 6.01`

Now let's find `f'(x)` as `deltax ->0`

`f'(x) = lim_(deltax->0) (f(x+deltax)-f(x))/(deltax)`
N.B. This is the definition of the 1st derivative of `f(x)`

Since, `f(x) = x^2`

`f'(x)= lim_(deltax->0) (x^2+2xdeltax+deltax^2 - x^2)/(deltax`

`= lim_(deltax->0) 2x+deltax = 2x`

So, the accurate value of `f'(3) = 2xx3 = 6`
Hence, our estimate above was out by `0.01`