Question

Over which interval does `f(t) = 2^t -t` have a negative average rate of change?

Answer 1

`t<-log_2(ln2)`

Explanation:

First differentiate - define `z = 2^x` for simplicity.

`:. z = 2^x`
`= (e^(ln2))^x`
`= e^(x*ln2)`

Define `u = x*ln2`

`:. z = e^u`
`:. (dz)/(du) = e^u`
(The derivative of `e` to the power of any number is itself.)

`(du)/(dx) = ln2`.

`:. dy/dx = e^(x*ln2)*ln2`
`= 2^x*ln2`.

Defining `f(x) = 2^t - t`

`:. f'(t) = 2^t*ln2 - 1`.

Letting `f'(t) < 0`

`:. 2^t*ln2 < 1`
`:. 2^t < 1/ln2`
`:. t < log_2(ln2)^-1`
`:. t < -log_2(ln2)`.

Or, `t< (-ln(ln2)) / ln2`.

Answer 2

There are infinitely many intervals over which `f` has a negative rate of change.

Explanation:

On any interval `[a,b}` with `f(a) > f(b)`, the average rate of change, `(f(b)-f(a))/(b-a)` is negative.

Pick an interval using the graph below.

graph{2^x-x [-6.063, 5.036, -0.603, 4.947]}

For example `[-4,2]` or `[-20,1}` or any of infinitely many intervals.