IV. Let
f
′
(
a
)
=
f
″
(
a
)
=
⋯
=
f
(
n
−
l
)
(
a
)
=
0
,
{\displaystyle f'(a)=f''(a)=\cdots =f^{(n-l)}(a)=0,}
and
f
(
n
)
(
a
)
≠
0
{\displaystyle f^{(n)}(a)\neq 0}
By continuing the process as illustrated in I, II, and III, it is seen that if the first derivative of
f
(
x
)
{\displaystyle f(x)}
x = a is of even order (= n ), then
(47)
f
(
a
)
{\displaystyle f(a)}
f
(
n
)
(
a
)
=
{\displaystyle f^{(n)}(a)=}
(48)
f
(
a
)
{\displaystyle f(a)}
f
(
n
)
(
a
)
=
{\displaystyle f^{(n)}(a)=}
[1]
If the first derivative of
f
(
x
)
{\displaystyle f(x)}
x = a is of odd order, then
f
(
a
)
{\displaystyle f(a)}
Illustrative Example 1.
x
3
−
9
x
2
+
24
x
−
7
{\displaystyle x^{3}-9x^{2}+24x-7}
Solution.
f
(
x
)
=
x
3
−
9
x
2
+
24
x
−
7
{\displaystyle f(x)=x^{3}-9x^{2}+24x-7}
f
′
(
x
)
=
3
x
2
−
18
x
+
24
{\displaystyle f'(x)=3x^{2}-18x+24}
Solving
3
x
2
−
18
x
+
24
=
0
{\displaystyle 3x^{2}-18x+24=0}
gives the critical values x = 2 and x = 4. ∴
f
′
(
2
)
=
0
{\displaystyle f'(2)=0}
f
′
(
4
)
=
0
{\displaystyle f'(4)=0}
Differentiating again,
f
″
(
x
)
=
6
x
−
18
{\displaystyle f''(x)=6x-18}
Since
f
″
(
2
)
=
−
6
{\displaystyle f''(2)=-6}
(47 ) that
f
(
2
)
=
13
{\displaystyle f(2)=13}
Since
f
″
(
4
)
=
+
6
{\displaystyle f''(4)=+6}
(48 ) that
f
(
4
)
=
9
{\displaystyle f(4)=9}
Illustrative Example 2.
e
x
+
2
cos
x
+
e
−
x
{\displaystyle e^{x}+2\cos x+e^{-x}}
Solution.
f
(
x
)
=
e
x
+
2
cos
x
+
e
−
x
{\displaystyle f(x)=e^{x}+2\cos x+e^{-x}}
f
′
(
x
)
=
e
x
−
2
sin
x
−
e
−
x
=
0
{\displaystyle f'(x)=e^{x}-2\sin x-e^{-x}=0}
x = 0,[2]
f
″
(
x
)
=
e
x
−
2
cos
x
+
e
−
x
=
0
{\displaystyle f''(x)=e^{x}-2\cos x+e^{-x}=0}
x = 0,
f
‴
(
x
)
=
e
x
+
2
sin
x
−
e
−
x
=
0
{\displaystyle f'''(x)=e^{x}+2\sin x-e^{-x}=0}
x = 0,
f
i
v
(
x
)
=
e
x
+
2
cos
x
+
e
−
x
=
4
{\displaystyle f^{iv}(x)=e^{x}+2\cos x+e^{-x}=4}
x = 0.
Hence from (48 ) ,
f
(
0
)
=
4
{\displaystyle f(0)=4}
EXAMPLES
Examine the following functions for maximum and minimum values, using the method of the last section:
1.
3
x
4
−
4
x
3
+
1
{\displaystyle 3x^{4}-4x^{3}+1}
Ans.
x = 1 gives min. = 0;
x = 0 gives neither.
2.
x
3
−
6
x
2
+
12
x
+
48
{\displaystyle x^{3}-6x^{2}+12x+48}
x = 2 gives neither.
3.
(
x
−
1
)
2
(
x
+
1
)
3
{\displaystyle (x-1)^{2}(x+1)^{3}}
x = 1 gives min. = 0 ;
x
=
1
5
{\displaystyle {\begin{smallmatrix}x={\frac {1}{5}}\end{smallmatrix}}}
x = -1 gives neither.
4. Investigate
x
6
−
5
x
4
+
5
x
3
−
1
{\displaystyle x^{6}-5x^{4}+5x^{3}-1}
x = 1 and x = 3.
5. Investigate
x
3
−
3
x
2
+
3
x
+
7
{\displaystyle x^{3}-3x^{2}+3x+7}
x = 1.
6. Show that if the first derivative of
f
(
x
)
{\displaystyle f(x)}
x = a is of odd order (= n ), then
f
(
x
)
{\displaystyle f(x)}
x = a , according as
f
(
n
)
(
a
)
{\displaystyle f^{(n)}(a)}
↑ As in § 82 , a critical value x = a is found by placing the first derivative equal to zero and solving the resulting equation for real roots.
↑ x = 0 is the only root of the equation
e
x
−
2
sin
x
−
e
−
x
=
0
{\displaystyle e^{x}-2\sin x-e^{-x}=0}
