Calculate range of the function:
f(x) = (x⁷+x⁵+x³-x²-1)/6
Answers
Answer:
Given : x + 1/x = 3
To find : value of x² + 1/x², x³ + 1/x³, and x⁴+1/x⁴
Solution :
We have x + 1/x = 3 ……..(1)
On squaring eq 1 both sides,
(x + 1/x)² = 3²
By Using Identity : (a + b)² = a² + b² + 2ab
x² + 1/x² + 2 x × 1/x = 9
x² + 1/x² + 2 = 9
x² + 1/x² = 9 - 2
x² + 1/x² = 7 ………….(2)
On squaring eq 2 both sides,
(x² +1/x² )² = 7²
(x²)² + (1/x²)² + 2 x² × 1/x² =7²
x⁴ + 1/x⁴ + 2 = 49
x⁴ + 1/x⁴ = 49 - 2
x⁴ + 1/x⁴ = 47
On Cubing eq 1 both sides :
(x + 1/x)³ = 3³
By Using Identity : (a + b)³ = a³ + b³ + 3ab(a + b)
(x)³ + (1/x)³ + 3 × x× 1/x (x + 1/x) = 27
x³ + 1/x³ + 3(x + 1/x) = 27
x³ + 1/x³ + 3(3) = 27
x³ + 1/x³ + 9 = 27
x³ + 1/x³ = 27 - 9
x³ + 1/x³ = 18
Hence the value of the value of x² + 1/x² is 7 , x³ + 1/x³ is 18 & x⁴ + 1/x⁴ is 47.
HOPE THIS ANSWER WILL HELP YOU…..Given : x + 1/x = 3
To find : value of x² + 1/x², x³ + 1/x³, and x⁴+1/x⁴
Solution :
We have x + 1/x = 3 ……..(1)
On squaring eq 1 both sides,
(x + 1/x)² = 3²
By Using Identity : (a + b)² = a² + b² + 2ab
x² + 1/x² + 2 x × 1/x = 9
x² + 1/x² + 2 = 9
x² + 1/x² = 9 - 2
x² + 1/x² = 7 ………….(2)
On squaring eq 2 both sides,
(x² +1/x² )² = 7²
(x²)² + (1/x²)² + 2 x² × 1/x² =7²
x⁴ + 1/x⁴ + 2 = 49
x⁴ + 1/x⁴ = 49 - 2
x⁴ + 1/x⁴ = 47
On Cubing eq 1 both sides :
(x + 1/x)³ = 3³
By Using Identity : (a + b)³ = a³ + b³ + 3ab(a + b)
(x)³ + (1/x)³ + 3 × x× 1/x (x + 1/x) = 27
x³ + 1/x³ + 3(x + 1/x) = 27
x³ + 1/x³ + 3(3) = 27
x³ + 1/x³ + 9 = 27
x³ + 1/x³ = 27 - 9
x³ + 1/x³ = 18
Hence the value of the value of x² + 1/x² is 7 , x³ + 1/x³ is 18 & x⁴ + 1/x⁴ is 47.
HOPE THIS ANSWER WILL HELP YOU…..Given : x + 1/x = 3
To find : value of x² + 1/x², x³ + 1/x³, and x⁴+1/x⁴
Solution :
We have x + 1/x = 3 ……..(1)
On squaring eq 1 both sides,
(x + 1/x)² = 3²
By Using Identity : (a + b)² = a² + b² + 2ab
x² + 1/x² + 2 x × 1/x = 9
x² + 1/x² + 2 = 9
x² + 1/x² = 9 - 2
x² + 1/x² = 7 ………….(2)
On squaring eq 2 both sides,
(x² +1/x² )² = 7²
(x²)² + (1/x²)² + 2 x² × 1/x² =7²
x⁴ + 1/x⁴ + 2 = 49
x⁴ + 1/x⁴ = 49 - 2
x⁴ + 1/x⁴ = 47
On Cubing eq 1 both sides :
(x + 1/x)³ = 3³
By Using Identity : (a + b)³ = a³ + b³ + 3ab(a + b)
(x)³ + (1/x)³ + 3 × x× 1/x (x + 1/x) = 27
x³ + 1/x³ + 3(x + 1/x) = 27
x³ + 1/x³ + 3(3) = 27
x³ + 1/x³ + 9 = 27
x³ + 1/x³ = 27 - 9
x³ + 1/x³ = 18
Hence the value of the value of x² + 1/x² is 7 , x³ + 1/x³ is 18 & x⁴ + 1/x⁴ is 47.
HOPE THIS ANSWER WILL HELP YOU…..Given : x + 1/x = 3
To find : value of x² + 1/x², x³ + 1/x³, and x⁴+1/x⁴
Solution :
We have x + 1/x = 3 ……..(1)
On squaring eq 1 both sides,
(x + 1/x)² = 3²
By Using Identity : (a + b)² = a² + b² + 2ab
x² + 1/x² + 2 x × 1/x = 9
x² + 1/x² + 2 = 9
x² + 1/x² = 9 - 2
x² + 1/x² = 7 ………….(2)
On squaring eq 2 both sides,
(x² +1/x² )² = 7²
(x²)² + (1/x²)² + 2 x² × 1/x² =7²
x⁴ + 1/x⁴ + 2 = 49
x⁴ + 1/x⁴ = 49 - 2
x⁴ + 1/x⁴ = 47
On Cubing eq 1 both sides :
(x + 1/x)³ = 3³
By Using Identity : (a + b)³ = a³ + b³ + 3ab(a + b)
(x)³ + (1/x)³ + 3 × x× 1/x (x + 1/x) = 27
x³ + 1/x³ + 3(x + 1/x) = 27
x³ + 1/x³ + 3(3) = 27
x³ + 1/x³ + 9 = 27
x³ + 1/x³ = 27 - 9
x³ + 1/x³ = 18
Hence the value of the value of x² + 1/x² is 7 , x³ + 1/x³ is 18 & x⁴ + 1/x⁴ is 47.
HOPE THIS ANSWER WILL HELP YOU :)