Question

How do you find the critical numbers of `y= 2x-tanx`?

Answer 1

`x=pi/4+(kpi)/2,pi/2+kpi,kinZZ`

Explanation:

Note that a critical number of a function `f` will occur at `x=a` when `f'(a)=0` or `f'(a)` is undefined.

So, we first need to find the derivative of the function.

`y=2x-tanx`

`dy/dx=2-sec^2x`

So, critical values will occur when:

`0=2-sec^2x`

Or:

`sec^2x=2" "=>" "secx=+-sqrt2" "=>" "cosx=+-1/sqrt2=+-sqrt2/2`

Note that this occurs at `x=pi/4,(3pi)/4,(5pi)/4,(7pi)/4...` which can be summarized as `x=pi/4+(kpi)/2` where `kinZZ`, which means that `k` is an integer.

Furthermore, note that `2-sec^2x` is undefined for some values:

`2-sec^2x=2-1/cos^2x`

This is undefined when `cosx=0`, which occurs at `x=pi/2,(3pi)/2,(5pi)/2...` or `x=pi/2+kpi,kinZZ`.

So, we have critical values at `x=pi/4+(kpi)/2,pi/2+kpi,kinZZ`.