Question

Use the formula for instantaneous rate of change, approximating the limit by using smaller and smaller values of hh , to find the instantaneous rate of change for each function at the given value.

f(x)=x^{x} at x=2

Answer 1

`Rate = 6.7726`

Explanation:

Given

`f(x) = x^x` at
`x =2`

Required

The instantaneous rate of change

We have:

`f(x) = x^x`

The instantaneous rate of change is:

`\lim_(h \to 0) (f(a + h) -f(a))/(h)`

`x =2` implies that:
`a = 2`

So, we have:

`a = 2`
`h = 0.01`

`(f(a + h) -f(a))/(h) = (f(2 + 0.01) -f(2))/(0.01) = (f(2.01) -f(2))/(0.01) = (2.01^(2.01) - 2^2)/(0.01) = 6.840403`

Keep reducing h but set a constant at 2

`h = 0.001`

`(f(a + h) -f(a))/(h) = (f(2 + 0.001) -f(2))/(0.001) = (f(2.001) -f(2))/(0.001) = (2.001^(2.001) - 2^2)/(0.001) = 6.779327`

`h = 0.0001`

`(f(a + h) -f(a))/(h) = (f(2 + 0.0001) -f(2))/(0.0001) = (f(2.0001) -f(2))/(0.0001) = (2.0001^(2.0001) - 2^2)/(0.0001) = 6.773262`

`h = 0.00001`

`(f(a + h) -f(a))/(h) = (f(2 + 0.00001) -f(2))/(0.00001) = (f(2.00001) -f(2))/(0.00001) = (2.00001^(2.00001) - 2^2)/(0.00001) = 6.772656`

`h = 0.000001`

`(f(a + h) -f(a))/(h) = (f(2 + 0.000001) -f(2))/(0.000001) = (f(2.000001) -f(2))/(0.000001) = (2.000001^(2.000001) - 2^2)/(0.000001) = 6.772595`

`h = 0.0000001`

`(f(a + h) -f(a))/(h) = (f(2 + 0.0000001) -f(2))/(0.0000001) = (f(2.0000001) -f(2))/(0.0000001) = (2.0000001^(2.0000001) - 2^2)/(0.0000001) = 6.772589`

`h = 0.00000001`

`(f(a + h) -f(a))/(h) = (f(2 + 0.00000001) -f(2))/(0.00000001) = (f(2.00000001) -f(2))/(0.00000001) = (2.00000001^(2.00000001) - 2^2)/(0.00000001) = 6.772589`

Notice that:

`(f(a + h) -f(a))/(h) = 6.772589` for
`h = 0.00000001` and
`h = 0.0000001`

Hence, the instantaneous rate of change is:

`Rate = 6.772589`

`Rate = 6.7726` ---- approximated