SME
Calculus I · Problem type
Sketching x·e−x/k from f′ and f″
Use this for “Sketch f(x) = 3x·e−x/2: where is it largest, and where does it change concavity?”
Chart
Local maximum at (2, ≈ 0.7358)
This is an example. Type your own values to start a new problem.
f(x)
=
A
x·
e
−x/
xc
Critical point
worked out (not typed)
f(xc)
Maximum value
worked out (not typed)
xᵢ
Inflection point
worked out (not typed)
f(xᵢ)
Value at the inflection
worked out (not typed)
Formulas
xc
=
k
2 = 2
f
(xc)
=
A
×
k
×
e
−1
0.7358
=
1
×
2
×
e
-1
xᵢ
=
2
×
k
4 = 2 × 2
f
(xᵢ)
=
2
×
A
×
k
×
e
−2
0.5413
=
2
×
1
×
2
×
e
-2
Assumptions
•
f(x) = Ax·e−x/k with A > 0 and k > 0. Its slope f′(x) = A·e−x/k(1 − x ÷ k) is positive for x < k and negative after, so f rises, then falls.
•
f″(x) = (A ÷ k)e−x/k(x ÷ k − 2) is negative before x = 2k and positive after: concave down, then concave up.
•
As x grows, e−x/k shrinks faster than x grows, so the curve tends to 0 without reaching it.
Step-by-step
WE KNOW
A = 1, k = 2
FIND
critical point (xc), maximum value (f(xc)), inflection point (xᵢ), value at the inflection (f(xᵢ))
Step 1 · Find critical point (xc)
Use xc = k
Set
f
′(x)
=
0.
The
factor
e
−x/
k
is
never
0,
so
1
−
x
k
=
0
and
x
=
k
.
The
slope
turns
from
+
to
−,
so
this
is
the
maximum.
f
(x)
=
x·
e
−
x
2
f
′(x)
=
e
−
x
2
−
(
1
2
)x·
e
−
x
2
=
e
−
x
2
(1
−
x
2
)
1
−
x
2
=
0
→
x
=
2
xc = 2
Step 2 · Find maximum value (f(xc))
Use f(xc) = A × k × e⁻¹
Put
x
=
k
into
f:
A
×
k
×
e
−
k
k
,
and
e
−1
=
1
e
.
f
(xc)
=
1
×
2
×
e
-1
f
(xc)
=
2
×
0.3679
f
(xc)
=
0.7358
Step 3 · Find inflection point (xᵢ)
Use xᵢ = 2 × k
Set
f
″(x)
=
0.
Only
the
last
factor
x
k
−
2
can
be
0,
at
x
=
2
k
,
and
f″
changes
sign
there,
so
the
bend
changes.
xᵢ = 2 × 2
f
″(x)
=
(
1
2
)
e
−
x
2
(
x
2
−
2)
x
2
−
2
=
0
→
x
=
4
xᵢ = 4
Step 4 · Find value at the inflection (f(xᵢ))
Use f(xᵢ) = 2 × A × k × e⁻²
Put
x
=
2
k
into
f:
A
×
2
k
×
e
−2
k
/
k
,
and
e
−2
=
1
e
2
.
f
(xᵢ)
=
2
×
1
×
2
×
e
-2
f
(xᵢ)
=
2
×
2
×
0.13534
f
(xᵢ)
=
4
×
0.13534
f
(xᵢ)
=
0.5413
Check
Put the numbers back in. Both sides should match.
f
′(2)
=
1
×
e
−1
×
(1
−
2
2
)
=
0
✓
0.7358
=
1
×
2
×
e
-1
✓
f
″(4)
=
(
1
2
)
×
e
−2
×
(
4
2
−
2)
=
0
✓
0.5413
=
2
×
1
×
2
×
e
-2
✓
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