SME
Calculus I
Mean value theorem and curve sketching
Use this for “Find every c in (1, 4) where f(x) = x³ − 3x² + 2x has the slope of its secant from 1 to 4.”
Chart
Tangent at x = 2.732: slope f′(2.732) = 8, y = 8x − 18.39
This is an example. Type your own values to start a new problem.
f(x)
=
x³
+
(
q
)
x²
+
r
x
f′(
c
)
=
f(
)
−
f(
)
−
=
m
f(a)
Value at the left end
worked out (not typed)
f(b)
Value at the right end
worked out (not typed)
Formulas
f
(
a
)
=
a
3
+
q
×
a
2
+
r
×
a
0
=
1
3
+
(−3)
×
1
2
+
2
×
1
f
(
b
)
=
b
3
+
q
×
b
2
+
r
×
b
24
=
4
3
+
(−3)
×
4
2
+
2
×
4
m
×
(
b
−
a
)
=
f
(
b
)
−
f
(
a
)
8 × (4 − 1) = 24 − 0
3
×
c
2
+
2
×
q
×
c
+
r
=
m
,
a
<
c
<
b
3
×
2.7321
2
+
2
×
(−3)
×
2.7321
+
2
=
8,
1
<
2.7321
<
4
Assumptions
•
f(x) = x³ + qx² + rx is continuous on [a, b] and differentiable on (a, b), as every polynomial is.
•
The mean value theorem then promises at least one c in (a, b) where the tangent’s slope f′(c) equals the secant’s slope m.
•
Rolle’s theorem is the case f(a) = f(b): the secant is level, m = 0, and the tangent at c is level too.
Step-by-step
WE KNOW
q = −3, r = 2, a = 1, b = 4
FIND
value at the left end (f(a)), value at the right end (f(b)), secant slope (m), point where f′(c) = m (c)
Step 1 · Find value at the left end (f(a))
Use f(a) = a³ + q × a² + r × a
Put x = a into f: the height where the secant starts.
f
(
a
)
=
1
3
+
(−3)
×
1
2
+
2
×
1
f
(
a
)
=
1
+
(−3)
+
2
f
(
a
)
=
0
Step 2 · Find value at the right end (f(b))
Use f(b) = b³ + q × b² + r × b
Put
x
=
b
into
f:
the
height
where
the
secant
ends.
f
(
b
)
=
4
3
+
(−3)
×
4
2
+
2
×
4
f
(
b
)
=
64
+
(−3)
×
16
+
2
×
4
f
(
b
)
=
64
+
(−48)
+
8
f
(
b
)
=
24
Step 3 · Find secant slope (m)
Use m × (b − a) = f(b) − f(a)
The
secant’s
slope
is
its
rise
f
(
b
)
−
f
(
a
)
over
its
run
b
−
a
.
m
=
f
(
b
) −
f
(
a
)
b
−
a
m
=
24 − 0
4 − 1
m
=
24
3
m
=
8
Step 4 · Find point where f′(c) = m (c)
Use 3 × c² + 2 × q × c + r = m, a < c < b
Set
f
′(
c
)
=
m
and
solve
the
quadratic
3
c
²
+
2qc
+
(
r
−
m
)
=
0.
Keep
the
root
that
lies
in
(
a
,
b
).
c
=
(−
q
+
√(
q
2
−
3
×
(
r
−
m
)))
÷
3
c
=
(−(−3)
+
√(
(−3)
2
−
3
×
(2
−
8)))
÷
3
f
(x)
=
x
3
−
3x²
+
2x
→
f
′(x)
=
3x²
−
6x
+
2
3
c
²
−
6
c
−
6
=
0
c
=
2.7321
or
c
=
−0.7321;
in
(1,
4):
c
=
2.7321
c
=
2.7321
Check
Put the numbers back in. Both sides should match.
0
=
1
3
+
(−3)
×
1
2
+
2
×
1
✓
24
=
4
3
+
(−3)
×
4
2
+
2
×
4
✓
8 × (4 − 1) = 24 − 0 ✓
3
×
2.7321
2
+
2
×
(−3)
×
2.7321
+
2
=
8,
1
<
2.7321
<
4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.