1. Solve graphically: y = x + 2 = 1 - x
Answers
Answer:
Step-by-step explanation:
Observe that if
x=±a, y=a21+a2 i.e. for every x,y), (−x,y
)
too lies on the curve. Hence, the graph is symmetric w.r.t.
y
-axis.
Now as
y
=
x
2
1
+
x
2
=
1
+
x
2
−
1
1
+
x
2
=
1
−
1
(
1
+
x
)
2
=
1
−
(
1
+
x
2
)
−
2
and hence,
d
y
d
x
=
−
(
−
2
(
1
+
x
2
)
3
×
2
x
)
=
4
x
(
1
+
x
2
)
3
Observe that as
y
=
1
−
1
(
1
+
x
)
2
, value of
y
or range of
x
2
1
+
x
2
is limited between
[
0
,
1
)
. Further, as
d
y
d
x
=
0
only at
x
=
0
and
x
=
±
∞
and hence we have an extrema at these points.
Now for second derivative using quotient formula, it is
d
2
y
d
x
2
=
(
1
+
x
2
)
3
×
4
−
4
x
×
3
(
1
+
x
2
)
2
×
2
x
(
1
+
x
2
)
6
=
(
1
+
x
2
)
2
[
4
+
4
x
2
−
24
x
2
)
(
1
+
x
2
)
6
=
4
(
1
−
5
x
2
)
(
1
+
x
2
)
4
and this is zero when
x
2
=
0.2
or
x
=
±
0.447
Alsso note that when
x
=
0
,
d
2
y
d
x
2
>
0
, hence wehave a minima at
x
=
0
and when
x
→
±
∞
,
d
2
y
d
x
2
→
0
, but on the negative side. Further, as
x
→
±
∞
,
y
→
1
and hence we have a maxima at
±
∞
.
The graph appears as follows:
graph{x^2/(1+x^2) [-5, 5, -1.62, 3.38]}