Sec 4.2 Applications of the Second
Derivative
Concave up at a point
= there is a tangent at that
point and f’ is increasing.
Concave down at a point
= there is a tangent at that
point and f’ is decreasing
= there is a tangent at that
point and f’’>0 (**)
=there is a tangent at that
point and f’’ < 0 (**)
= the graph lies above the
tangent at that point.
= the graph lies below the
tangent at that point
(**) it may so happen that the graph is concave up or concave
down at a point and f’’(x)=0 at that point. So if f’’=0 at a
point you have to make some additional calculation to see how the
value of f’ changes as we move across the point to determine the
concavity of the graph at that point. If f’ is increasing then it
is concave up, if it is decreasing then it is concave down.
The graph is concave up in (− ∞,− 1.5) ⋃ (0.7, 2.7)
The graph is concave down in (− 1.5, 0.7) ⋃ (2, 7, ∞)
It has inflection points at (-1.5,f(-1.5)) , (0.7,f(0.7)) and
(2,7,f(2.7))
Inflection point.​ A point on the graph, where ​there is a tangent
and where the ​concavity of the graph changes​ is called an
inflection point.
** The inflection point is also known as the point of diminishing returns.
. if graph is concave up at a point
then the g​ raph lies above the tangent
. if the graph is concave down at a
point the the g​ raph lies below the
tangent
. at an inflection point t​ he tangent
crosses the graph.
(**) f’’=0 does not tell you anything about the concavity. It may
be concave up, it may be concave down or it may be an inflection
point.
How to determine the intervals in which the graph is concave up
and the intervals in which it is concave down, and the inflection
points.
1. Find all the point where the f ′′(x) = 0 or f ′′(x) does not
exist.
2. Find the sign diagram of f ′′(x)
3. Find the sign of f ′′(x) in each of the intervals. Depending
upon the sign of f ′′(x) determine the intervals of positive
and negative concavity.
4. (​for inflection points​) Choose among the points found in
step 1, those where the graph has a tangent( that is f’(x)
exists or it has a vertical tangent). For each of these
points from the the sign diagram determine if the sign of
f’’ changes as we move across the point. If so, then this
point is an inflexion point. If not, it is not an inflection
point.
(**) The function must be defined at an inflection point and the
graph must have a tangent at that point. You have to make sure of
this when using any point found in step 1.
Q. For the function f (x) = x + 1x determine the intervals where the
graph of the function is concave up and the intervals where it is
concave down.
Soln. f ′(x) = 1 − x12 . f ′′(x) = x23
f ′′(x) is never 0, and it is undefined at x=0
So the ​sign diagram of f ′′(x) ​ is
-------------------------dne+++++++++++++++++++++++
f ′′(− 1) = − 2
f ′′(1) = 2
0
The function is concave down in (− ∞, 0)
The function is concave up in (0, ∞)
The only candidate for an inflection point is x=0, but this can’t
be an inflection point since the function is not defined there(
a.k.a there is no point on the graph corresponding to x=0). ​So
this graph does not have any inflection points.
Q. ​Determine the intervals where the graph of the function
f (x) = x21+1 is concave up, and the intervals where it is concave
down. And find the inflection points of f (x) .
Soln. f ′(x) = −
4
2
2x
(x2+1)2
+2]
f ′′(x) = − [−6x(x2−4x
=
+1)4
= −
2x
x4+2x2+1
6x4+4x2−2
(x2+1)4
=
. f ′′(x) = −
[ 2(x4+2x2+1) − 2x(4x3+4x)]
(x2+1)4
(6x2−2)(x2+1)
(x2+1)4
f ′′(x) exists everywhere, and it is zero when 6x2 − 2 = 0, x2 + 1 = 0
The first equation gives x = √13 , x = √−13 . the second equation does
not have any solutions. So the sign diagram for f ′′(x) is :
+++++++++++++++++++0-------------------0+++++++++++++++++
−1
f ′′(0) =
√3
up in (− ∞, √−13 ) ⋃ ( √13 , ∞)
down in ( √−13 , √13 )
f ′′(2) = 22/25
Concave
Concave
−2
1
√3
f ′′(2) = 22/25
Since f ′′(x) exists at both x =± √13 , at both these point the
concavity changes, so both ( √−13 , 3/4) ,( √13 , 3/4) are inflection
points.