SME
Calculus I · Problem type
Highs and lows of a cubic
Use this for “Find the local maximum and minimum of f(x) = x³ − 6x² + 9x + 1, and where the graph changes how it bends.”
Chart
Local maximum at (−1, 10)
Local minimum at (3, −22)
This is an example. Type your own values to start a new problem.
f(x)
=
x³
+
(
b
)
x²
+
(
c
)
x
+
d
x₁
First critical point
worked out (not typed)
k₁
Second derivative f″(x₁) there
worked out (not typed)
y₁
Local maximum f(x₁)
worked out (not typed)
x₂
Second critical point
worked out (not typed)
k₂
Second derivative f″(x₂) there
worked out (not typed)
y₂
Local minimum f(x₂)
worked out (not typed)
xᵢ
Inflection point
worked out (not typed)
yᵢ
Value at the inflection f(xᵢ)
worked out (not typed)
Formulas
x₁
=
−
b
−
√
b
2
− 3 ×
c
3
−1
=
(−(−3)
−
√(
(−3)
2
−
3
×
(−9)))
÷
3
x₂
=
−
b
+
√
b
2
− 3 ×
c
3
3
=
(−(−3)
+
√(
(−3)
2
−
3
×
(−9)))
÷
3
k₁
=
6
×
x₁
+
2
×
b
−12 = 6 × (−1) + 2 × (−3)
k₂
=
6
×
x₂
+
2
×
b
12 = 6 × 3 + 2 × (−3)
y₁
=
x₁
3
+
b
×
x₁
2
+
c
×
x₁
+
d
10
=
(−1)
3
+
(−3)
×
(−1)
2
+
(−9)
×
(−1)
+
5
y₂
=
x₂
3
+
b
×
x₂
2
+
c
×
x₂
+
d
−22
=
3
3
+
(−3)
×
3
2
+
(−9)
×
3
+
5
xᵢ
=
−
b
3
1
=
−
−3
3
yᵢ
=
xᵢ³
+
b
×
xᵢ²
+
c
×
xᵢ
+
d
−6
=
1
3
+
(−3)
×
1
2
+
(−9)
×
1
+
5
Assumptions
•
The local maximum and minimum are at critical points, where f′(x) = 3x² + 2bx + c = 0.
•
Two critical points need b² > 3c. Then the second derivative test sorts them: f″(x) = 6x + 2b is negative at a maximum and positive at a minimum.
•
With x³ in front, the smaller critical point x₁ is always the maximum and x₂ the minimum. The inflection point is halfway, where f″(x) = 0.
Step-by-step
WE KNOW
b = −3, c = −9, d = 5
FIND
first critical point (x₁), second critical point (x₂), second derivative f″(x₁) there (k₁), second derivative f″(x₂) there (k₂), local maximum f(x₁) (y₁), local minimum f(x₂) (y₂), inflection point (xᵢ), value at the inflection f(xᵢ) (yᵢ)
Step 1 · Find first critical point (x₁)
Use x₁ = (−b − √(b² − 3 × c)) ÷ 3
Set
f
′(x)
=
3x²
+
2bx
+
c
=
0.
The
quadratic
formula
gives
x
=
−2
b
±
√
4
b
² − 12
c
6
;
cancel
the
2s.
The
smaller
root
is
x₁
.
x₁
=
(−(−3)
−
√(
(−3)
2
−
3
×
(−9)))
÷
3
x₁
=
(−(−3)
−
√(9
−
3
×
(−9)))
÷
3
x₁
=
(−(−3)
−
√(9
−
(−27)))
÷
3
x₁
=
−(−3) −
√
36
3
x₁
=
3 −
√
36
3
x₁
=
3 − 6
3
x₁
=
−
3
3
x₁
=
−1
Step 2 · Find second critical point (x₂)
Use x₂ = (−b + √(b² − 3 × c)) ÷ 3
The
same
formula
with
+
gives
the
larger
root,
x₂
.
x₂
=
(−(−3)
+
√(
(−3)
2
−
3
×
(−9)))
÷
3
x₂
=
(−(−3)
+
√(9
−
3
×
(−9)))
÷
3
x₂
=
(−(−3)
+
√(9
−
(−27)))
÷
3
x₂
=
−(−3) +
√
36
3
x₂
=
3 +
√
36
3
x₂
=
3 + 6
3
x₂
=
9
3
x₂
=
3
Step 3 · Find second derivative f″(x₁) there (k₁)
Use k₁ = 6 × x₁ + 2 × b
Differentiate
twice
with
the
power
rule.
f
″(
x₁
)
<
0,
so
the
graph
bends
down
there:
x₁
is
a
local
maximum.
k₁
=
6
×
(−1)
+
2
×
(−3)
f
(x)
=
x
3
−
3x²
−
9x
+
5
→
f
′(x)
=
3x²
−
6x
−
9
→
f
″(x)
=
6x
−
6
f
″(−1)
=
6(−1)
−
6
=
−12
k₁
=
−12
Step 4 · Find second derivative f″(x₂) there (k₂)
Use k₂ = 6 × x₂ + 2 × b
f
″(
x₂
)
>
0,
so
the
graph
bends
up
there:
x₂
is
a
local
minimum.
k₂
=
6
×
3
+
2
×
(−3)
f
″(3)
=
6(3)
−
6
=
12
k₂
=
12
Step 5 · Find local maximum f(x₁): y₁
Use y₁ = x₁³ + b × x₁² + c × x₁ + d
Put
x₁
into
f:
the
height
of
the
local
maximum.
y₁
=
(−1)
3
+
(−3)
×
(−1)
2
+
(−9)
×
(−1)
+
5
y₁
=
−1
+
(−3)
+
(−9)
×
(−1)
+
5
y₁
=
−1
+
(−3)
+
9
+
5
y₁
=
10
Step 6 · Find local minimum f(x₂): y₂
Use y₂ = x₂³ + b × x₂² + c × x₂ + d
Put
x₂
into
f:
the
height
of
the
local
minimum.
y₂
=
3
3
+
(−3)
×
3
2
+
(−9)
×
3
+
5
y₂
=
27
+
(−3)
×
9
+
(−9)
×
3
+
5
y₂
=
27
+
(−27)
+
(−27)
+
5
y₂
=
−27
+
5
y₂
=
−22
Step 7 · Find inflection point (xᵢ)
Use xᵢ = −b ÷ 3
Set
f
″(x)
=
6x
+
2
b
=
0.
On
the
left
f″
<
0
and
on
the
right
f″
>
0,
so
the
bend
changes
from
down
to
up
here.
xᵢ
=
−
−3
3
xᵢ = 1
Step 8 · Find value at the inflection f(xᵢ): yᵢ
Use yᵢ = xᵢ³ + b × xᵢ² + c × xᵢ + d
Put
xᵢ
into
f.
A
cubic
is
symmetric
about
this
point,
so
yᵢ
is
halfway
between
y₁
and
y₂
.
yᵢ
=
1
3
+
(−3)
×
1
2
+
(−9)
×
1
+
5
yᵢ = 1 + (−3) + (−9) + 5
yᵢ = −6
Check
Put the numbers back in. Both sides should match.
−1
=
(−(−3)
−
√(
(−3)
2
−
3
×
(−9)))
÷
3
✓
3
=
(−(−3)
+
√(
(−3)
2
−
3
×
(−9)))
÷
3
✓
−12 = 6 × (−1) + 2 × (−3) ✓
12 = 6 × 3 + 2 × (−3) ✓
10
=
(−1)
3
+
(−3)
×
(−1)
2
+
(−9)
×
(−1)
+
5
✓
−22
=
3
3
+
(−3)
×
3
2
+
(−9)
×
3
+
5
✓
1
=
−
−3
3
✓
−6
=
1
3
+
(−3)
×
1
2
+
(−9)
×
1
+
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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