Determine where, if anywhere, the tangent line to f(x) = x3 – 5x² + x has a slope of -2.
Answers
Step-by-step explanation:
Firstly, we'll compare the given equation of the parallel line with the general slope-intercept form
y
=
m
x
+
c
for the value of the slope m of the tangent line.
The differentiation of the given cubic curve at a point is known as the slope of the tangent line so we'll equate the obtained derivative equal to the value of the slope and simplify the obtained equation for the variable x.
Answer and Explanation: 1
Given information:
The given cubic polynomila curve is:
f
(
x
)
=
x
3
−
5
x
2
+
x
The equation of the line that is parallel to a tangent line is:
y
=
4
x
+
23
Comparing the linear equation or parallel equation
y
=
4
x
+
23
with the general slope-intercept form
y
=
m
x
+
c
, we get:
m
=
4
Differentiating the given cubic curve with respect to the variable x, we get:
d
f
(
x
)
d
x
=
d
(
x
3
−
5
x
2
+
x
)
d
x
f
′
(
x
)
=
(
3
x
3
−
1
−
5
(
2
)
x
2
−
1
+
1
)
f
′
(
x
)
=
3
x
2
−
10
x
1
+
1
=
3
x
2
−
10
x
+
1
Equating the slope of the given parallel line equal to the above expression, we get:
f
′
(
x
)
=
m
3
x
2
−
10
x
+
1
=
4
3
x
2
−
10
x
+
1
−
4
=
0
3
x
2
−
10
x
−
3
=
0
Simplifying the above quadratic equation for the values of the variable x, we get:
x
=
−
(
−
10
)
±
√
(
−
10
)
2
−
4
(
3
)
(
−
3
)
2
(
3
)
x
=
10
±
√
100
+
36
6
x
=
10
±
2
√
34
6
x
=
5
±
√
34
3
=
5
+
√
34
3
,
5
−
√
34
3
=
3.61
,
−
0.28
By the given curve
y
=
f
(
x
)
=
x
3
−
5
x
2
+
x
at
x
=
3.61
, the value of the variable y is:
y
=
(
3.61
)
3
−
5
(
3.61
)
2
+
3.61
=
47.045881
−
65.1605
+
3.61
=
−
14.50
The value of the variable y at
x
=
−
0.28
is:
y
=
(
−
0.28
)
3
−
5
(
−
0.28
)
2
+
(
−
0.28
)
=
−
0.021952
−
0.392
−
0.28
=
−
0.69
Thus, the points where a tangent of the given curve is parallel to the given line are
(
3.61
,
−
14.50
)
and
(
−
0.28
,
−
0.69
)
.