Which of the following functions f has a removable discontinuity at a? If the discontinuity is removable, find a function g that agrees with f for x≠a and is continuous at a. f(x)=x4−1x−1 ,a=1

Question

Which of the following functions f has a removable discontinuity at a? If the discontinuity is removable, find a function g that agrees with f for x≠a and is continuous at a. f(x)=x4−1x−1 ,a=1

Sact1989 · Accepted Answer

Given the function  f(x)=x4−1x−1  we want to see whether it has a removable discontinuity. Observe that the function f is not defined at x=1 and hence function is not continuous at x=1. Now in order se whether it is removable or not, let us find out the limit.  limx→1f(x)=limx→1x4−1x−1  =limx→1(x2−1)(x2+1)x−1  =limx→1(x−1)(x+1)(x2+1)x−1  =limx→1(x2+1)(x+1)  =limx→1(x2+1)⋅limx→1(x+1)  =(1+1)(1+1)  =4  Since the limit exists at x=1 and therefore x=1 is a removable discontinuity.  Define the function  g(x)={f(x),if x≠14,if x=1  Then the function g is continuous at 1 as  limx→1g(x)=limx→1f(x)=4=g(1)

user_27qwe · Accepted Answer

Step 1:To find the limit, we can factor the numerator of f(x) as a difference of squares:f(x)=x4−1x−1=(x2−1)(x2+1)x−1.Canceling the common factor of x−1, we have:f(x)=(x2−1)(x2+1)x−1=(x−1)(x+1)(x2+1)x−1=(x+1)(x2+1).Thus, we see that f(x) simplifies to (x+1)(x2+1) for x≠1.Step 2:To find a function g that agrees with f for x≠1 and is continuous at x=1, we can simply replace f(x) with g(x)=(x+1)(x2+1):g(x)=(x+1)(x2+1).The function g(x) is continuous at x=1 since there are no longer any factors in the denominator that could cause a discontinuity.Therefore, the function f(x)=x4−1x−1 has a removable discontinuity at x=1, and a function g(x)=(x+1)(x2+1) can be defined to be equal to f(x) for x≠1 and continuous at x=1.

star233 · Accepted Answer

Answer:g(x)={x4−1x−1,if&amp;nbsp;x≠14,if&amp;nbsp;x=1Explanation:Let&#039;s start by analyzing the function f(x)=x4−1x−1.To find the behavior of f as x approaches 1, we can factor the numerator using the difference of squares formula: x4−1=(x2+1)(x2−1). This simplifies the function to:f(x)=(x2+1)(x2−1)x−1Now, we can cancel out the common factor of x−1 in the numerator and denominator:f(x)=(x2+1)(x+1)(x−1)x−1Simplifying further, we get:f(x)=(x2+1)(x+1)Notice that f(x) is defined for all x except when x=1, where the denominator becomes 0. This suggests a potential discontinuity at x=1.To confirm whether the discontinuity is removable or not, we need to evaluate the limit of f(x) as x approaches 1. Let&#039;s compute the limit:limx→1f(x)=limx→1(x2+1)(x+1)Since (x2+1)(x+1) is a polynomial function, the limit can be found by directly substituting x=1:limx→1(x2+1)(x+1)=(12+1)(1+1)=2·2=4The limit exists and is finite, indicating that the discontinuity at x=1 is removable.To define a function g that agrees with f for x≠1 and is continuous at 1, we can redefine g(x) as g(x)=f(x) for x≠1 and g(1)=4 (the value of the limit we found).Thus, the function g(x) that satisfies these conditions is:g(x)={x4−1x−1,if&amp;nbsp;x≠14,if&amp;nbsp;x=1The function g(x) is continuous at x=1 because limx→1g(x)=g(1)=4.In summary, the function f(x)=x4−1x−1 has a removable discontinuity at a=1. The function g(x) that agrees with f(x) for x≠1 and is continuous at x=1 is given by:g(x)={x4−1x−1,if&amp;nbsp;x≠14,if&amp;nbsp;x=1

alenahelenash · Accepted Answer

To determine if the function f has a removable discontinuity at a=1, we need to check if the limit of f exists as x approaches a and if it is finite.Let&#039;s calculate the limit of f as x approaches 1:limx→1f(x)=limx→1x4−1x−1Since this is an indeterminate form of 00, we can apply L&#039;Hôpital&#039;s rule by differentiating the numerator and the denominator with respect to x:limx→14x31=4The limit exists and is finite, which means the discontinuity at x=1 is removable.To find a function g that agrees with f for x≠1 and is continuous at x=1, we can simply define g(x) to be equal to the limit of f(x) as x approaches 1. Therefore, we have:g(x)=limx→1f(x)=4Hence, the function g(x) that agrees with f for x≠1 and is continuous at x=1 is given by g(x)=4.

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