Determine the set of points at which the function is continuous. f(x,y)=(x2y32x2+y2) if (x,y) not equal to (0,0) and 1 if (x,y)=(0,0)

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pseudoenergy34 · Accepted Answer

On 𝟚R2 without the origin, f is a rational function, and so is continuous at every point in its domain. Since it is defined on this entire domain, we see that f is continuous at every point.  To check whether f is continuous at the origin, we must check whether  lim(x,y)⇒(0,0)f(x,y)=f(0,0).  Checking along the line y=0, we see that  limx⇒0x2(0)32x2+(0)2=limx⇒00=0  But this implies that if f has a limit as (x,y) tends to (0,0), it must be 0, which is not the value of f(0,0). Thus we see that in this case, f(x,y) is not continuous at the origin.  Result:  f(x,y) is not continuous at the origin.

nick1337 · Accepted Answer

Answer:{(x,y)∈ℝ2:(x,y)≠(0,0)}Explanation:To determine the set of points at which the function is continuous, we need to examine the behavior of the function at different points.The given function is:f(x,y)={x2y32x2+y2,if&amp;nbsp;(x,y)≠(0,0)1,if&amp;nbsp;(x,y)=(0,0)Let&#039;s first analyze the behavior of the function when (x,y)≠(0,0). In this case, the function is defined as:f(x,y)=x2y32x2+y2We notice that the function is a rational function, which is continuous everywhere except for the points where the denominator is equal to zero. Hence, we need to determine when 2x2+y2=0.Since 2x2 is always non-negative, for the sum 2x2+y2 to be zero, both 2x2 and y2 must be zero. This implies x=0 and y=0. However, this contradicts the assumption that (x,y)≠(0,0).Therefore, for all points (x,y)≠(0,0), the function f(x,y) is continuous.Now, let&#039;s analyze the behavior of the function at the point (0,0). In this case, the function is defined as:f(x,y)=1Since the function is defined as a constant, it is continuous at (0,0).In summary, the function f(x,y) is continuous at all points except (0,0). The set of points at which the function is continuous is given by:{(x,y)∈ℝ2:(x,y)≠(0,0)}

Eliza Beth13 · Accepted Answer

Step 1:To determine the set of points at which the function is continuous, let&#039;s analyze the function f(x,y)=x2y32x2+y2 for (x,y)≠(0,0) and 1 for (x,y)=(0,0).First, we need to check the continuity of the function at (0,0). To do this, we&#039;ll examine the limit of the function as (x,y) approaches (0,0).lim(x,y)→(0,0)f(x,y)=lim(x,y)→(0,0)x2y32x2+y2To evaluate this limit, we can use polar coordinates. Let&#039;s substitute x=rcosθ and y=rsinθ into the function:limr→0(rcosθ)2(rsinθ)32(rcosθ)2+(rsinθ)2Simplifying this expression gives:limr→0r5cos2θsin3θ2r2cos2θ+r2sin2θNow, we can cancel out the r2 term from the denominator:limr→0r3cos2θsin3θ2cos2θ+sin2θStep 2:Next, we can take the limit as r approaches 0:limr→0r3cos2θsin3θ=0Thus, the limit of the function as (x,y) approaches (0,0) exists and is equal to 0. Therefore, the function is continuous at (0,0).Now, let&#039;s consider the continuity of the function for (x,y)≠(0,0). Since the function is a rational function, it is continuous everywhere except where the denominator is equal to 0.Setting the denominator 2x2+y2 equal to 0, we have:2x2+y2=0Since both x2 and y2 are non-negative, the equation above cannot hold for any real values of x and y. Thus, the denominator is never equal to 0 for (x,y)≠(0,0).Therefore, the function is continuous for all (x,y)∈ℝ2.

Don Sumner · Accepted Answer

To determine the set of points at which the function is continuous, we need to examine the behavior of the function as we approach different points. Let&#039;s analyze the function f(x,y)=x2y32x2+y2 when (x,y) is not equal to (0,0), and f(x,y)=1 when (x,y)=(0,0).First, let&#039;s investigate the function&#039;s behavior as (x,y) approaches (0,0). We&#039;ll consider approaching along different paths.Approaching along the x-axis, we let y=0. Substituting into the function, we have:limx→0f(x,0)=limx→0x2·032x2+02=limx→002x2=0Approaching along the y-axis, we let x=0. Substituting into the function, we have:limy→0f(0,y)=limy→002·y32·02+y2=limy→00y2=0Thus, the function approaches 0 as we approach (0,0) along the x-axis and the y-axis. To determine if the function is continuous at (0,0), we need to check if the limit of the function as (x,y) approaches (0,0) exists and is equal to the value of the function at (0,0).Let&#039;s compute the limit of the function as (x,y) approaches (0,0) along the path y=mx, where m is a constant:lim(x,y)→(0,0)f(x,y)=limx→0x2(mx)32x2+(mx)2=limx→0m3x52x2+m2x2=limx→0m3x5(2+m2)x2=m3·02+m2=0Since the limit is 0 for all paths, the limit exists and is equal to the value of the function at (0,0), which is 1. Therefore, f(x,y) is continuous at (0,0).In conclusion, the function f(x,y)=x2y32x2+y2 is continuous for all points (x,y) in the entire xy-plane, including (0,0).

Determine the set of points at which the function is

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