Question

Determine where this function is continuous

Answer 1

notice the junction points
see if the function's overlap each other at the "junctions"

Answer 2

ummm, they don't overlap?

Answer 3

@jdoe0001 ?

Answer 4

they don't? dunno, have you checked?

Answer 5

let's see the function that's to the left-hand side \(\bf 20-x \qquad x \le -5\)
what's the value of that at the "junction" of -5?

Answer 6

25?

Answer 7

so, the other function that "takes over" from there is \(x^2 \qquad -5 < x < 2\)
what's the value of that one at the junction of -5?

Answer 8

25 ! so those overlap!

Answer 9

low and behold!, they overlap each other, so there's continuity from \(\bf 20-x \ and \ x^2\)

now let's check the other "junction point", 2
what's \(\bf x^2\) at 2?

Answer 10

4

Answer 11

ok, from 2, the next function that takes over there, to the right-hand side is \(\bf 2+x \qquad x \ge 2\)

what's the value of that one at 2?

Answer 12

4 ! so \[x ^{2} \] and x+2 overlap!

Answer 13

or 2+x i mean. same thing

Answer 14

yes they do overlap, so a the "junction" points they overlap each other
the 1st one is a line, the 2nd one is a parabola, the 3rd one is a line

the functions themselves are continuous, and if they overlap at those junction points, that means?

Answer 15

so in interval notation...hold on give me a sec

Answer 16

keep in mind that 20-x goes to \(\bf -\infty\)
and that 2+x goes to \(\bf +\infty\)

Answer 17

\[(-\infty,-5)U(-5,2)U(2,\infty)\]

Answer 18

?

Answer 19

@jdoe0001

Answer 20

\(\bf (-\infty,-5)\cup(-5,2)\cup(2,\infty) \implies (-\infty\quad ,\quad +\infty)\)

Answer 21

oooooh i see.

Answer 22

that was very helpful. thank you.

Answer 23

the piecewise function, as it jumps from one to the next, is continuous all around :)