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Graph the function and observe where it is discontinuous.

f(x, y) = e¹/⁽ˣ⁻ʸ⁾
f is discontinuous at x =

User Eddiemoya
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Final answer:

To graph the function f(x, y) = e^(1/(x-y)), we find it is discontinuous where x = y, as the function is not defined when the denominator of the exponent is zero. Other examples like the horizontal line f(x) = 10/20 on the interval 0 ≤ x ≤ 20 are continuous, while the graph of f(x) = 1/x has an asymptote at x = 0.

Step-by-step explanation:

When graphing the function f(x, y) = e1/(x-y), we can identify points of discontinuity by looking for values of x and y where the function is not defined or cannot produce a real number output. In the context of this function, discontinuity occurs if the denominator in the exponent, (x - y), is zero because division by zero is undefined. Therefore, the function f(x, y) is discontinuous where x = y.

However, in the provided examples, such as a graph of a function f(x) on the interval 0 ≤ x ≤ 20 where f(x) is a horizontal line and f(x) = 10/20, we are dealing with a function of a single variable which is continuous in its domain. Similarly, the graph of f(x) = 1/x shows a common type of discontinuity called an asymptote at x = 0, since f(x) approaches infinity as x gets closer to zero.

User Rushby
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