Find the value of the constant 𝑎 that will make this piecewise function continuous everywhere.𝑓(𝑥)=⎧⎩⎨⎪⎪𝑥+1𝑎1−𝑥𝑥<0𝑥=0𝑥>0

The function is given as:

f(x) = { x + 1 if x < 0 a if x = 0 1 - x if x > 0 }

For the function to be continuous everywhere, the limit of the function as x approaches 0 from the left (x < 0) must be equal to the limit of the function as x approaches 0 from the right (x > 0), and both must be equal to the value of the function at x = 0.

Let's find these values:

1. Limit as x approaches 0 from the left (x < 0):

lim (x->0-) (x + 1) = 0 + 1 = 1

2. Limit as x approaches 0 from the right (x > 0):

lim (x->0+) (1 - x) = 1 - 0 = 1

3. Value of the function at x = 0:

f(0) = a

For the function to be continuous at x = 0, these three values must be equal:

1 = 1 = a

So, the value of the constant a that will make this piecewise function continuous everywhere is a = 1.