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Given f(x)=kx+1 for x<1 and f(x)=x^2 for x `>=` 1 ` `
(a) Find k so that f is continuous at x=1:
A function is continuous at a point if it is defined at the point ( f(c) exists), the limit at the point exists, and the limit agrees with the defined value of the function at the point.
At x=1, the value of the function is 1. Thus in order to be continuous at x=1, the limit from the left and the right must agree and be 1.
Let k=0. Then f(1)=1 (since 1^2=1); the limit from the right is 1 and the limit from the left is 1. Thus the function is continuous at x=1.
(b) F is continuous if it is continuous at every point. With k=0, for x<1 the function is x=1 which is continuous everywhere. For x>1 the function is f(x)=x^2 which is continuous everywhere. And the function is continuous at x=1. Thus the function is continuous.
(c) Let g(x) be 2x-3 for x not equal to 2, and let g(x)=0 for x=2. This function has a removable discontinuity at x=2.
If we redefine g(x) so that g(x)=2x-3 for x not equal to 2, and g(x)=1 for x=2 the function is continuous.
My teacher explained that if you are given a piece-wise function and asked if the function is continuous, you need 3 pieces of evidence: is the function defined, does the limit exist, and does the limit match the function value. Basically, everything has to equal.
You will know if the function is defined based on if there is an equals sign or not. In your problem, you can see that x is greater than or equal to 1 for the function x^2. Therefore, you can say the function f(x) is defined at x=1 because x^2=(1)^2=1. To find if the limit exists, you must insert 1 into both of the equations given to see if they are equal (you already did that for x^2 and the limit from the right is 1). If you want the function to be continuous, your limit from the left has to equal 1. This would mean k=0. kx+1=(0)(1)+1=1. This makes the function f(x) continuous at x=1 because the limit as x approaches 1 equals the function value. This is your part A.
Part B, simply wants you to explain why it is continuous using k. The function is continuous when k=0 because there are no gaps in the function. Each number is defined when x is less than 1, equal to 1, and greater than 1.
In part c, the function has a removable discontinuity at x=2, because neither function is equal to 2. For the function to be continuous there has to be a defined spot at x=2. This can be fixed by simply changing the 0 in the bottom function for another number, such as 1. Now, g(x)=1 at x=2 and the function is continuous.
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