Second derivative of a function is used to determine the concavity, convexity, the points of inflection, and local extrema of functions.

**Applications of Second
Derivative**

Second
derivative of a function is used to determine the concavity, convexity, the
points of inflection, and local extrema of functions.

** **

A graph
is said to be concave down (convex up) at a point if the tangent line lies
above the graph in the vicinity of the point. It is said to be concave up (convex
down) at a point if the tangent line to the graph at that point lies below the
graph in the vicinity of the point. This may be easily observed from the
adjoining graph.

Let *f* ( *x*) be a function whose second derivative
exists in an open interval *I* = (*a*,
*b*) . Then the function *f* ( *x*)
is said to be

(i) If *f* ′( *x*)
is strictly increasing on *I* , then
the function is concave up on an open interval *I*

(ii) If *f* ′( *x*)
is strictly decreasing on *I* , then
the function is concave down on an open interval *I*.

Analytically,
given a differentiable function whose graph *y*
=
*f* ( *x*) , then the concavity is given by the following result.

(i) If *f* ′′(*x*) > 0 on an open interval *I* , then *f* ( *x*) is concave up on *I* .

(ii) If *f* ′′(*x*) < 0 on an open interval *I* , then *f* ( *x*) is concave down on *I* .

**Remark**

(1) Any
local maximum of a convex upward function defined on the interval [*a* , *b*]
is also its absolute maximum on this interval.

(2) Any
local minimum of a convex downward function defined on the interval [*a* , *b*]
is also its absolute minimum on this interval.

(3) There
is only one absolute maximum (and one absolute minimum) but there can be more
than one local maximum or minimum.

** **

The
points where the graph of the function changes from **“concave up to concave down”** or **“concave down
to concave up” **are called the** points of
inflection of ***f*** **(** ***x*)** **.

(i) If *f *′′(*c*)* *exists and* f *′′(*c*)* *changes sign when
passing through* x *=* c *, then the point* *(*c*
, *f* (*c*)) is a point of inflection of the graph of *f* .

(ii) If *f* ′′(*c*)
exists at the point of inflection, then *f*
′′(*c*) = 0 .

**Remark**

To
determine the position of points of inflexion on the curve *y* = *f* ( *x*) it is necessary to find the points
where *f* ′′(
*x*) changes sign. For ‘smooth’ curves
(no sharp corners), this may happen when either

(i) *f *′′(*x*)* *=* *0*
*or

(ii) *f *′′(*
x*)* *does not exist at the point.

**Remark**

(1) It
is also possible that *f* ′′(*c*) may not exist, but (*c* , *f*
(*c*)) could be a point of inflection.
For instance, *f* (*x*) = *x*^{1/3}
at *c* = 0 .

(2) It
is possible that *f* ′′(*c*) = 0 at a point but (*c* , *f*
(*c*)) need not be a point of inflection.
For instance, *f* (*x*) = *x*^{4}
at *c* = 0 .

(3) A
point of inflection need not be a stationary point. For instance, if *f* (*x*)
=
sin *x* then, *f *′* (x)* = cos* x *and *f *′′*(x)* = − sin* x *and hence (π, 0) is a point of inflection but not a stationary point
for *f (x)* .

**Example 7.57**

Determine
the intervals of concavity of the curve *f*
(*x*) = (*x* −1)^{3} ⋅ ( *x* − 5), *x* ∈ ℝ and, points of inflection if any.

**Solution**

The
given function is a polynomial of degree 4. Now,

*f* ′(*x*) = *(x *− 1)^{3} + 3*(x *− 1)^{2} ⋅ (x − 5)

= 4(x −
1)^{2} ⋅
(x − 4)

*f* ′′ (*x*)= 4((x − 1)^{2}
+ 2(x − 1) ⋅ (x −
4))

= 12(x − 1) ⋅ (x − 3)

Now,

*f* ′′( x) = 0 ⇒* x *=1,* x *= 3 .

The
intervals of concavity are tabulated in Table 7.7.

The
curve is concave upwards on (−∞,1) and (3, ∞)
.

The
curve is concave downwards on (1, 3) .

As *f* ′′(*x*)
changes its sign when it passes through *x*
= 1 and *x* = 3, (1,* f *(1)) = (1, 0) and (3, *f* (3)) = (3, −16) are points of inflection for the
graph *y* =
*f* ( *x*) . The sign change may be observed from the adjoining figure of
the curve *f* ′′(
*x*) .

** **

Determine
the intervals of concavity of the curve *y*
=
3 +
sin *x* .

**Solution**

The
given function is a periodic function with period 2*π*
and hence there will be stationary points and points of inflections in each
period interval. We have,

We now
consider an interval, (−*π*
, *π* ) by splitting into two sub intervals
(−π , 0) and ( 0, π) .

In the
interval (−π , 0) , d^{2}*y/dx*^{2}* *> 0 and hence the function is
concave upward.

In the
interval (0,π ), d^{2}*y/dx*^{2}* *< 0 and hence the function is concave
downward. Therefore (0, 3) is a point of inflection (see Fig. 7.25). The
general intervals need to be considered to discuss the concavity of the curve
are (*n**π*
, (*n* +1)*π*
) , where *n* is any integer which can
be discussed as before to conclude that (*n**π*, 3) are also points of inflection.

** **

**The Second Derivative Test: **The Second Derivative Test relates
the concepts of critical points,** **extreme
values, and concavity to give a very useful tool for determining whether a
critical point on the graph of a function is a relative minimum or maximum.

Suppose
that *c* is a critical point at which *f* ′(*c*)
=
0 , that *f* ′(
*x*) exists in a neighborhood of *c* , and that* f *′′(c) exists. Then* f *has
a relative maximum value at c if* f *′′
(c) < 0 and a relative minimum value at c if* f *′′* *(c ) > 0 . If *f* ′′ (c) = 0 , the test is not
informative.

**Example 7.59**

Find the
local extremum of the function *f* (*x*) = *x*^{4}
+
32*x* .

**Solution**

We have,

*f *′(*
x*)* *= 4*x*^{3} + 32 = 0 gives *x*^{3} = −8

*x *=* *−2

and* f *′′( x) = 12x^{2}.

As* f*′′*
*(−2) > 0, the function has local minimum at* x *= −2 . The local minimum value is *f* (− 2) = −48 . Therefore, the extreme point is (−2, −48).

Find the
local extrema of the function *f* (
*x* ) = 4*x*^{6} −6*x*^{4}
.

Differentiating
with respect to x, we get

* f* ′( *x*) _{=} 24*x*^{5} -24*x*^{3}

= 24*x*^{3}* *(* x*^{2}* *−1)

=
24*x*^{3} (
*x* +1)( *x*
−1)

* f *′(* x *)* *=* *0* *⇒* x *= −1,
0, 1. Hence the critical numbers are *x*
= −1,
0,1.

Now, *f* ′′( *x*
)
=
120*x*^{4} −72*x*^{2} =
24*x*^{2} (5
*x*^{2} −3).

⇒ *f *′′(−1)* *=* *48* *,*
f *′′(* *0)* *=* *0* *,* f *′′(1)* *=* *48* *.

As *f* ′′(−1) and *f* ′′(1) are positive by the second
derivative test, the function *f* (
*x*) has local minimum. But at *x* = 0, *f* ′′(0)
=
0 . That is the second derivative test does not give any information about
local extrema at *x* =
0 . Therefore, we need to go back to the first derivative test. The intervals
of monotonicity is tabulated in Table 7.8.

By the
first derivative test *f* ( *x*) has local minimum at *x* = −1, its local minimum value is −2
. At *x* = 0 , the function *f *(*
x*)* *has local maximum at* x *=* *0*
*, and its local maximum value is* *0.
At *x* = 1, the function *f* ( *x*)
has local minimum at *x* =
1, and its local minimum value is −2 .

When the
second derivative vanishes, we have no information about extrema. We have used
the first derivative test to find out the extrema of the function!

** **

Find the
local maximum and minimum of the function *x*
^{2} *y*^{2 }on the
line *x* + *y*
=
10 .^{}

**Solution**

Let the
given function be written as *f* (*x*) = *x*^{2}
(10 −
*x*)^{2}. Now,

*f (x)* =* x*^{2} (100 − 20*x*
+* x*^{2} ) = *x*^{4} − 20*x*^{3} +100*x*^{2}

Therefore,
*f ′(x)* = 4*x*^{3} − 60*x*^{2}
+ 200*x* = 4*x*(*x*^{2} −15x +
50)

* f ′(x)*
= 4*x*
(*x*^{2} − 15*x* + 50) = 0 ⇒* x *= 0, 5, 10

and *f*′′*
(x)* = 12*x*^{2} − 120*x* + 200

The
stationary numbers of *f *(* x*)*
*are* x *=* *0, 5, 10* *at these
points the values of *f *′′ (* x*)*
*are respectively 200, −100 and 200 . At *x* = 0 , it has local minimum and its
value is *f* (0) =
0 . At *x* =
5 , it has local maximum and its value is *f*
(5) =
625 . At *x* =
10 , it has local minimum and its value is *f*
(10) =
0 .

Tags : Applications of Differential Calculus | Mathematics , 12th Maths : UNIT 7 : Applications of Differential Calculus

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12th Maths : UNIT 7 : Applications of Differential Calculus : Applications of Second Derivative | Applications of Differential Calculus | Mathematics

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