Answer:
raph of function f(x) is concave down when f"(x) < 0
�
=
�
�
−
2
�
y=xe
−2x
Differentiating with respect to x
�
′
=
�
−
2
�
+
�
⋅
�
−
2
�
(
−
2
)
y
′
=e
−2x
+x⋅e
−2x
(−2)
⇒
�
′
=
�
−
2
�
(
1
−
2
�
)
⇒y
′
=e
−2x
(1−2x)
Differentiating with respect to x again
�
′
′
=
�
−
2
�
(
0
−
2
)
+
(
1
−
2
�
)
�
−
2
�
⋅
(
−
2
)
y
′′
=e
−2x
(0−2)+(1−2x)e
−2x
⋅(−2)
⇒
�
′
′
=
�
−
2
�
(
−
2
−
2
(
1
−
2
�
)
)
⇒y
′′
=e
−2x
(−2−2(1−2x))
⇒
�
′
′
=
�
−
2
�
(
−
2
−
2
+
4
�
)
⇒y
′′
=e
−2x
(−2−2+4x)
⇒
�
′
′
=
(
4
�
−
4
)
�
−
2
�
⇒y
′′
=(4x−4)e
−2x
point of inflection when
�
′
′
=
0
y
′′
=0
Since
�
−
2
�
e
−2x
is always positive
(
4
�
−
4
)
=
0
(4x−4)=0
�
=
1
x=1
We need to check sign of
�
′
′
y
′′
for x < 1 and x >1
Explanation:
�
′
′
=
�
−
2
⋅
0
(
4
⋅
0
−
4
)
=
1
(
0
−
4
)
=
−
4
<
0
y
′′
=e
−2⋅0
(4⋅0−4)=1(0−4)=−4<0 ................concave downward
when x > 1 (let x=2)
�
′
′
=
�
−
2
⋅
2
(
4
⋅
2
−
4
)
=
�
−
4
⋅
4
≈
0.07
>
0
y
′′
=e
−2⋅2
(4⋅2−4)=e
−4
⋅4≈0.07>0 ............concave upward
Thus, graph of
�
=
�
�
−
2
�
y=xe
−2x
is concave down for x <1
In interval notation, graph of
�
=
�
�
−
2
�
y=xe
−2x
is concave down in
(
−
∞
,
1
)
(−∞,1