Which of the following statements about the function f(x)=sec(x) are true? Select all that apply.
a. f(x) has no zeroes.
b. f(x) has a period of 2pi.*****
c. f(x) has an amplitude of 1.
d. f(x) is symmetric about the origin.
e. f(x) has asymptotes at integer multiples of pi.
I know it has a period of 2pi and I know that it has no amplitude. Other than that I do not know as there are two answers and I only know one. Any help is greatly appreciated. Thank you!!!
Well, you're right about the period of 2π! But, let's have some fun debunking the rest of these statements, shall we?
a. f(x) has no zeroes? Well, let's think about it. The secant function is equal to 1 divided by the cosine function. We know that the cosine function has zeroes at integer multiples of π. And since dividing by zero is a no-no, that means f(x) has zeroes at those points! So, statement a is false.
c. f(x) has an amplitude of 1? Nope, not even close. The amplitude of a function is a measure of its maximum value from the midline. The secant function doesn't have a maximum value because it keeps on going up to infinity and down to negative infinity. So, statement c is false.
d. f(x) is symmetric about the origin? Whoa, hold your horses there! The secant function is not symmetric about the origin. If you reflect it over the y-axis, it won't look the same! So, statement d is false.
e. f(x) has asymptotes at integer multiples of π? Absolutely! As I mentioned earlier, the secant function is equal to 1 divided by the cosine function. And we know that the cosine function has vertical asymptotes at integer multiples of π. So, statement e is true!
So, to recap: statements b and e are the ones that hold true. Keep in mind that math can be a bit tricky, but with a little humor, we can make it more enjoyable!
Based on the information provided, the answer is:
b. f(x) has a period of 2pi.
Explanation:
a. f(x) has no zeroes: This statement is not true. The function f(x) = sec(x) has zeroes when x takes on the values of pi/2 + npi, where n is an integer.
c. f(x) has an amplitude of 1: This statement is not true. The function f(x) = sec(x) does not have an amplitude since it does not represent any kind of oscillation or periodic movement.
d. f(x) is symmetric about the origin: This statement is not true. The function f(x) = sec(x) is not symmetric about the origin. It is, however, symmetric about the vertical lines x = pi/2 and x = 3pi/2.
e. f(x) has asymptotes at integer multiples of pi: This statement is true. The function f(x) = sec(x) has vertical asymptotes at x = npi, where n is an integer.
To determine which statements about the function f(x) = sec(x) are true, let's go through each option one by one and explain how to find the correct answers:
a. f(x) has no zeroes:
To find the zeroes of a function, we need to solve the equation f(x) = 0. In this case, the function is sec(x). However, since sec(x) is equal to 1/cos(x), it does not equal zero at any value of x. Therefore, the statement that f(x) has no zeroes is true.
b. f(x) has a period of 2pi:
The period of a trigonometric function is the distance between two successive peaks or troughs of the function. For sec(x), the period is determined by the reciprocal of the period of the cosine function, which is 2pi. Hence, the statement that f(x) has a period of 2pi is true.
c. f(x) has an amplitude of 1:
The amplitude of a trigonometric function is the distance from the horizontal axis to the maximum or minimum value of the function. However, sec(x) does not have a maximum or minimum value in the traditional sense, as it oscillates between positive and negative infinity. Therefore, the statement that f(x) has an amplitude of 1 is false.
d. f(x) is symmetric about the origin:
A function is symmetric about the origin if it satisfies the property that f(-x) = -f(x) for all x in its domain. If we substitute -x into the function f(x) = sec(x), we get sec(-x), which is equal to 1/cos(-x). However, 1/cos(-x) is not equal to -sec(x), so the statement that f(x) is symmetric about the origin is false.
e. f(x) has asymptotes at integer multiples of pi:
An asymptote is a line that the graph of a function approaches but never touches. For sec(x), it has vertical asymptotes at all x-values where the cosine function, cos(x), is equal to zero. The cosine function equals zero at odd multiples of pi, which means sec(x) has vertical asymptotes at integer multiples of pi. Therefore, the statement that f(x) has asymptotes at integer multiples of pi is true.
In summary, the true statements are:
b. f(x) has a period of 2pi.
e. f(x) has asymptotes at integer multiples of pi.
sec x = 1/cos x
cos is between -1 and +1
when x = pi/2 or 3pi/2 the denominator is zero, boom
so it has no zeros
cos is symmetric about origin, so this is as well