1.Locate the discontinuity of the function
y=1/(1+sin x)
2.Suppose that f is a continuous function defined for all real numbers x and f(-5)=3 and f(-1)=-2 . If f(x) = 0 for one and only one value of x, then which of the following could be x?
#1: when is the denominator zero? When
1+sinx = 0
sinx = -1
x = 3π/2 + 2kπ
Since f(-5) > 0 and f(-1) < 0, f(x)=0 only for a single value of x between -5 and -1.
I assume the given choices included such a number.
1. To find the discontinuity of the function y=1/(1+sin x), we need to look for values of x where the denominator becomes zero.
The function y=1/(1+sin x) is undefined when the denominator, 1+sin x, equals zero.
Setting 1+sin x = 0, we can solve for x as follows:
sin x = -1
Solving for x, we find that x can be:
x = (2n - 1)π, where n is an integer.
So, the discontinuity of the function y=1/(1+sin x) occurs at x = (2n - 1)π, where n is an integer.
2. Since f is a continuous function defined for all real numbers x, it means that f(x) is defined for every value of x.
Given that f(-5)=3 and f(-1)=-2, we know that f(x) is continuous between x=-5 and x=-1, and the graph of f(x) must intersect the x-axis at least once within this interval.
For f(x) to equal zero for one and only one value of x, it means that the graph of f(x) must intersect the x-axis at exactly one point.
Therefore, the only possibility for x is a single value within the interval (-5, -1), and any value within this interval could be the value of x where f(x)=0.
1. To locate the discontinuity of the function y = 1/(1 + sin x), we need to identify the points where the function is not defined or where it exhibits a jump or hole.
The function y = 1/(1 + sin x) is not defined when the denominator (1 + sin x) equals zero. Solving for sin x = -1, we find that sin x = -1 when x = (2n - 1)π - π/2, where n is an integer.
Therefore, the function is discontinuous at x = (2n - 1)π - π/2 for all integer values of n. These are the points where the denominator becomes zero, resulting in a discontinuity.
2. If f is a continuous function defined for all real numbers x and f(-5) = 3 and f(-1) = -2, we can use the Intermediate Value Theorem to determine which values of x satisfy f(x) = 0.
The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b], and f(a) < 0 and f(b) > 0 (or vice versa), then there exists at least one number c in the interval (a, b) such that f(c) = 0.
In this case, f(-5) = 3 > 0 and f(-1) = -2 < 0. Therefore, by the Intermediate Value Theorem, there exists at least one value of x between -5 and -1 such that f(x) = 0.
However, without further information about the behavior of the function f, we cannot determine which specific value of x makes f(x) = 0. Additional information, such as the shape of the graph or the properties of the function, would be needed to find the exact value of x.