Answer:
f(x) = x2
For critical
points f?(x) = 0
Now f?(0) = 2
has local
minima at x = 0 and local minimum value of f(x) is f(0) = 0.
II. g(x)
= x3 ?3x
Now
g??(x) = 6x
g?(x)|x=
?1 = ?6 < 0
g?(x)|x=1
= 6 > 0
g(x) has
local maxima at x = ?1
and
local minimum at x = 1
Local
max. value = g(?1) = ?1 + 3 = 2
Local
min. value = g(1) = 1?3 = ?2
III. h(x)
= sin x + cos x in
h?(x)
= 0
Now
h?(x) = ? (sin x + cos x)
h?(x)
= ? (sin x + cos x)
has local
maxima at and local
max. value
IV. f(x)
= sin x ? cos x in (0, )
f?(x)
= 0
Now
has local
maxima at
and
local minima at
Local max.
value
and
local min. value =
V. f(x)
= x3 ? 6x2 + 9x + 15
Now
f?(x) = 6x ? 12 = 6(x ? 2)
f?(x)|x=1
= ?6 < 0
f?(x)|x
= 3 = 6 > 0
f(x) has
local maxima at x = 1
and
local minima at x = 3
Local
max. Value = f(1)
=
1 ? 6 + 9 + 15 = 19
and
local minimum value = f(3)
=
27 ? 54 + 27 + 15 = 15
VI.
g?(x)
= 0
Since
x > 0, therefore taking x = 2
Now
g?(x) = 0 +
has local
minima at x = 2 and local min.
VII.
g?(x)
= 0
Now
g? (x) = ? 2
has local
maxima at x = 0 and local maximum value = g(0) =
VIII.
f?(x)
= 0
Now
f?(x) =
has local
maxima at
Local
maximum value
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