Find the critical numbers of the functions.
f(0) = 2sec0 +tan0
on the interval [0,2pi]
0 = theta
why not just use x?
f(x) = 2secx+tanx
f' = 2secx tanx + sec^2x
= sec^2(x) (2tanx/secx+1)
= secx^2(x) (2sinx+1)
critical numbers are where
sec^2(x) = 0 -- no solutions there
sinx = -1/2 -- x = 7π/6, 11π/6
To find the critical numbers of a function, we need to find the values of x where the derivative of the function is equal to zero or does not exist.
First, let's express the function in terms of trigonometric functions only. Recall that sec(x) is equal to 1/cos(x) and tan(x) is equal to sin(x)/cos(x). Therefore, we can rewrite the function as:
f(x) = 2sec(x) + tan(x)
= 2(1/cos(x)) + sin(x)/cos(x)
= (2 + sin(x))/cos(x)
Now, we need to find the derivative of f(x) with respect to x. Applying the quotient rule, we have:
f'(x) = [(cos(x)(0) - (2 + sin(x))(-sin(x)))]/(cos^2(x))
= (-sin(x)(2 + sin(x)))/(cos^2(x))
= -sin(x)(2 + sin(x))/(cos^2(x))
To find the critical numbers, we set f'(x) equal to zero and solve for x:
-sin(x)(2 + sin(x))/(cos^2(x)) = 0
This equation is satisfied when either sin(x) = 0 or 2 + sin(x) = 0.
1. When sin(x) = 0, x can take the values of 0, pi, and 2pi.
2. When 2 + sin(x) = 0, we can solve for x:
sin(x) = -2
This equation has no solution since the range of sin(x) is [-1, 1].
Therefore, the critical numbers of the function f(x) = 2sec(x) + tan(x) on the interval [0, 2pi] are x = 0, pi, and 2pi.
To find the critical numbers of a function, we need to find the values of x where the derivative of the function is either zero or undefined. However, in this case, we are given a function f(x), not an explicit equation. Therefore, we need to find the derivative of the function first.
Let's differentiate the function f(x) = 2sec(x) + tan(x) with respect to x.
To find the derivative, we can use the quotient rule. The quotient rule states that if we have a function f(x) = g(x)/h(x), then the derivative is given by (g'(x)h(x) - g(x)h'(x))/(h(x))^2.
Applying the quotient rule, we get:
f'(x) = [2(sec(x))'tan(x) - 2sec(x)(tan(x))']/(sec(x))^2
To differentiate sec(x), we can use the chain rule:
(sec(x))' = sec(x)tan(x)
To differentiate tan(x), we can use the derivative of the tan function, which is sec^2(x).
Substituting these values into the derivative formula, we get:
f'(x) = [2(sec(x)tan(x))tan(x) - 2sec(x)(sec^2(x))]/(sec(x))^2
= [2sec(x)tan^2(x) - 2sec^3(x)]/(sec(x))^2
= [2sec(x)(tan^2(x) - sec^2(x))]/(sec(x))^2
Simplifying further:
f'(x) = [2sec(x)(tan^2(x) - sec^2(x))] / (sec(x))^2
= [2sec(x)(tan^2(x) - 1)] / (sec(x))^2
Now, to find the critical numbers of the function, we need to find the values of x where f'(x) = 0 or f'(x) is undefined.
Setting the numerator equal to zero, we have:
2sec(x)(tan^2(x) - 1) = 0
This equation is satisfied when sec(x) = 0 or tan^2(x) - 1 = 0.
First, let's solve sec(x) = 0:
sec(x) = 0 is equivalent to cos(x) = 1/0, which is undefined. Therefore, there are no critical numbers resulting from this condition.
Now, let's solve tan^2(x) - 1 = 0:
tan^2(x) - 1 = 0
tan^2(x) = 1
tan(x) = ±1
The solutions to this equation occur at x = π/4, 3π/4, 5π/4, and 7π/4, which are the critical numbers of the function f(x) = 2sec(x) + tan(x) on the interval [0, 2π].
To summarize, the critical numbers of the function f(x) = 2sec(x) + tan(x) on the interval [0, 2π] are x = π/4, 3π/4, 5π/4, and 7π/4.