find the point of inflexion on the curve r (theta^2-1) =a theta^2
Answers
Answer:
None, it's concave up for
Step-by-step explanation:
Compute the first derivative.
y
'
=
2
x
e
x
2
Second derivative is given by the product rule.
y
'
'
=
2
(
e
x
2
)
+
2
x
(
2
x
)
e
x
2
y
'
'
=
2
e
x
2
+
4
x
2
e
x
2
We need to set this to
0
and solve to determine inflection points.
0
=
2
e
x
2
+
4
x
2
e
x
2
0
=
2
e
x
2
(
1
+
2
x
2
)
We see this has no solution because
2
e
x
2
≠
0
for all values of
x
. Furthermore, the second equation states that
1
+
2
x
2
=
0
→
2
x
2
=
−
1
→
x
=
√
−
1
2
This doesn't have a real value so no real solution to this equation. This simply means the function
y
=
e
x
2
will have no inflection points (
y
'
'
is positive on all it's domain, therefore concave up on
(
−
∞
,
∞
)
). We can even confirm graphically.
enter image source here
Hopefully this helps!