Find the value of c for which the function is continuous for everywhere:

Question

f(x)=(c^2)+(x^2), x<1

f(x)=(c-x)^2,x\ge1

Answer 1

To find the value of c where f(x) is continuous everywhere, we need to know if the two parts of the function, f(x)=(c^2)+(x^2) and f(x)=(c-x)^2, meet at any point of potential discontinuity.

This is a piece-wise function, so f(x) is in two pieces. In this case, the only potential point of discontinuity is x=1, the parts where the two pieces meet. If the two parts of the function have the same value at x=1, then both parts will touch, and the function will be continuous at that point.

So, we substitute 1 in for x, and get c^2+1 and (c-1)^2

To make sure these are equal, we set them equal to each other and solve for c

c^2+1 = (c-1)^2

c^2+1 = c^2-2c+1

1=-2c+1

0=-2c

c=0

If c = 0, the two parts of this piece-wise function connect at x=1, which was the only potential point of discontinuity. So there you have it, c = 0

Hope this helps!