SME
Calculus I · Problem type
Make a piecewise function continuous
Use this for “Find k so that f(x) = 3x + k (x < 2), x² + 1 (x ≥ 2) is continuous.”
Chart
As x → 2 from the left, f(x) → 5; from the right, f(x) → 5.
So the limit is 5, equal to f(2): continuous there.
This is an example. Type your own values to start a new problem.
f(x)
=
m
x
+
(
k
)
for
x
<
c
f(x)
=
x²
+
d
for
x
≥
c
f(
c
)
=
f(c)
L⁻
Limit from the left at c
calculated
Formulas
L⁻
=
m
×
c
+
k
5 = 3 × 2 + (−1)
f
(
c
)
=
c
2
+
d
5
=
2
2
+
1
L⁻
=
f
(
c
)
5 = 5
Assumptions
•
f(x) = mx + k for x < c, and x² + d for x ≥ c.
•
f is continuous at c when the limit from the left equals f(c); the right piece already holds at c.
•
Each piece is continuous on its own, so x = c is the only place to check.
Step-by-step
WE KNOW
c = 2, m = 3, d = 1
FIND
value of f at c (f(c)), limit from the left at c (L⁻), constant of the left piece (k)
Step 1 · Find value of f at c (f(c))
Use f(c) = c² + d
Put
x
=
c
into
the
right
piece,
which
holds
at
c
.
f
(
c
)
=
2
2
+
1
f
(
c
)
=
5
Step 2 · Find limit from the left at c (L⁻)
Use L⁻ = f(c)
Continuous
at
c
:
the
left
limit
must
equal
f
(
c
).
L⁻ = 5
Step 3 · Find constant of the left piece (k)
Use L⁻ = m × c + k
Subtract
m
×
c
from
the
left
limit.
k
=
5
−
3
×
2
k
=
−1
Check
Put the numbers back in. Both sides should match.
5 = 3 × 2 + (−1) ✓
5
=
2
2
+
1
✓
5 = 5 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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Limits and continuity
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