Based on two different cases:
x
=
π
6
,
5
or
3
2
Look below for the explanation of these two cases.
Explanation:
Since,
cos
+
sin
1
we have:
−
So we can replace
in the equation
by
(
)
⇒
or,
0
using the quadratic formula:
b
±
√
4
a
c
for quadratic equation
⋅
8
9
Case I:
for the condition:
≤
to get positive value of
Case II:
to get negative value of
Answer link
Объяснение:
y = (x + 13)² * (e^x) - 15
Находим первую производную:
y` = (x + 13)² * (e^x) + (2x + 26) * (e^x) = (x + 13)*(x + 15) * (e^x)
Приравняем её к нулю:
(x + 13)*(x + 15) * (e^x) = 0
x₁ = - 13
x₂ = - 15
e^x > 0
Вычисляем значение функции:
f(-13) = - 15
f(- 15) = - 15 + 4/e¹⁵
fmin = - 15
fmax = - 15 + 4/e¹⁵
Используем достаточное условие экстремума функции для одной переменной.
y`` = (x + 13)² + 2*(2x + 26) * (e^x) + 2*(e^x) = (x² + 30x + 223) * (e^x)
Вычисляем:
y``(-15) = - 2/e¹⁵ < 0, значит эта точка - точка максимума
y``(-13) = 2/у¹³ > 0, значит эта точка - точка минимума
Based on two different cases:
x
=
π
6
,
5
π
6
or
3
π
2
Look below for the explanation of these two cases.
Explanation:
Since,
cos
x
+
sin
2
x
=
1
we have:
cos
2
x
=
1
−
sin
2
x
So we can replace
cos
2
x
in the equation
1
+
sin
x
=
2
cos
2
x
by
(
1
−
sin
2
x
)
⇒
2
(
1
−
sin
2
x
)
=
sin
x
+
1
or,
2
−
2
sin
2
x
=
sin
x
+
1
or,
0
=
2
sin
2
x
+
sin
x
+
1
−
2
or,
2
sin
2
x
+
sin
x
−
1
=
0
using the quadratic formula:
x
=
−
b
±
√
b
2
−
4
a
c
2
a
for quadratic equation
a
x
2
+
b
x
+
c
=
0
we have:
sin
x
=
−
1
±
√
1
2
−
4
⋅
2
⋅
(
−
1
)
2
⋅
2
or,
sin
x
=
−
1
±
√
1
+
8
4
or,
sin
x
=
−
1
±
√
9
4
or,
sin
x
=
−
1
±
3
4
or,
sin
x
=
−
1
+
3
4
,
−
1
−
3
4
or,
sin
x
=
1
2
,
−
1
Case I:
sin
x
=
1
2
for the condition:
0
≤
x
≤
2
π
we have:
x
=
π
6
or
5
π
6
to get positive value of
sin
x
Case II:
sin
x
=
−
1
we have:
x
=
3
π
2
to get negative value of
sin
x
Answer link
Объяснение: