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

Answer 1

Ezra Adams

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

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#.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer 2

Evelyn Agard

To find the critical numbers of $y = 2x - \tan(x)$ , we need to find where the derivative is either zero or undefined. First, find the derivative of the function $y$ , then set it equal to zero and solve for $x$ . Additionally, identify any points where the derivative is undefined.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.