Split domain function

Question

Is the split-domain function

f(x) = { x ^2 + 1 x<3

10 x = 3

(4/3)x + 6 x>3

continuous at x = 3?

Is it continuous on ℝ?

Answer 1

Hi Rachel,

Thanks for your question!

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For a function to be continuous at x = a,

f(a) = limit of f(x) as x --> a.

The limit of f(x) as x --> a exists if and only if

limit of f(x) as x --> a from the left = limit of f(x) as x --> a from the right.

Therefore, for a function to be continuous at x = a, we must have

f(a) = limit of f(x) as x --> a from the left = limit of f(x) as x --> a from the right.

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In this problem, f(x) is the piecewise function above and a = 3. To check continuity at x = a = 3, we must check whether

f(3) = limit of f(x) as x --> 3 from the left = limit of f(x) as x --> 3 from the right

is true or not. Since f(x) = x^2 + 1 to the left of x = 3, f(x) = 10 at x = 3, and f(x) = 4/3*x + 6 to the right of x = 3, we have to check whether

10 = limit of x^2 + 1 as x --> 3 = limit of 4/3*x + 6 as x --> 3

is true or not. We can evaluate both limits in this chain of equalities to find that, indeed,

10 = 10 = 10,

so we can be confident that our function is continuous at x = a = 3. Since x^2 + 1 and 4/3*x+6 are continuous functions for all x (they are polynomials), we have that our piece-wise function is continuous not just at x = 3, but the entire real line.

Answer 2

x^2 +1 replace x by 3 3^2+1 = 9+1 =10

10 with x=3 is still 10

4x/3 + 6 with x=3 is 4(3)/3+6 = 4+6 =10

as long as all 3 pieces = 10 at x =3, the function is continuous at x=3

f(x) is also continuous at all real values of x, since each of the 3 pieces are continuous

f(x) = x^2+1 is continuous, it's a parabola with no discontinuities, no denominator or square root problems

f(x) = 10 is horizontal line, with no discontinuous point, no denominator or square root problem

f(x) = (4/3)x +6 is a straight upward sloping line, again with no discontinuity, no denominator or square root problems

and the pieces meet, connect at x=3

major reason for a discontinuity within each piece is if some x value made a denominator =0 or a square root have an imaginary solution