Example of odd polynomial
Introduction:
Odd polynomial is nothing but a polynomial function with the odd numbers as powers. The general form of polynomial function is given as ax 2 + bx + c = 0. The odd polynomial in the sense the variables in the given functions have the power as odd terms. The general form of odd polynomial function is given as
ax 5 + b x 3 + cx + d = 0
The above function is named as odd polynomial function. The power terms present in the variable x are odd numbers. The following are the example problems for odd polynomials.
Example problems on odd polynomials:
Example 1:
Solve the odd polynomial function.
f(u) = u5 – 3u 3 – 4 u + u
Solution:
The given function is
f(u) = u5 – 3u 3 – 4 u + u
Differentiate the above odd polynomial function, we get
f '(u) = 5 u 4 – 3(3u 2 ) – 4( 1 ) + 1
f '(u) = 5u 4 – 9u 2 – 4 + 1
f '(u) = 5u 4 – 9u 2 – 3 is the solution.
Example 2:
Solve the odd polynomial function.
∫f(u) = u5+u3+18u + 12 du
Solution:
The given odd polynomial function is
∫f(u) = u5+u3+18u +12 du
∫f(u) = ∫ (u5+u3+18u +12) du
∫f(u) = ∫u5 du +∫u3 du+ ∫ 18u du +∫ 12
Integrate the above odd polynomial function, we get
F(u) =u6/6 + u4/4 + 18 u2 /2 + 12u
Simplifying the above function we get
F(u) =u6 / 6+ u4 / 4 +9u2 + 12u is the solution.
Example 3:
Solve the odd Polynomial Functions Examples.
∫f(u) = 4u3+3u+10 du
Solution:
The given odd polynomial function is
∫f(u) = 4u3+3u+10 du
∫f(u) = ∫ (4u3+3u+10) du
∫f(u) = ∫4u3 du +∫3u du+ ∫ 10
Integrate the above odd polynomial function, we get
F(u) =4u4/4 + 3u2/2 + 10 u
Simplifying the above function we get
F(u) =u4 + 3u2 /2+10u is the solution.
Example 4:
Solve the odd polynomial function.
∫f(u) = 4u3+3u+10u du
Solution:
The given odd polynomial function is
∫f(u) = 4u3+3u+10u du
∫f(u) = ∫ (4u3+3u+10u) du
∫f(u) = ∫4u3 du +∫3u du+ ∫ 10u du
Integrate the above odd polynomial function.
We get
F (u) =4u4/4 + 3u2/2 + 10 u2 /2
Simplifying the above function we get
F(u) =u4+ 3u2 / 2 +5u2
F(u) =u4+ 13u2 / 2 is the solution.
Practice problems on odd polynomials:
1) Solve the odd polynomial function.
f(x) = 4x 3 – 4 x + 5
Answer: f '(x) = 12x 2 + 4 is the solution.
2) Solve the odd polynomial function.
∫f(x) = x5+10x dx
Answer: F(x) =x6 / 6+ 5x2 is the solution.
Odd polynomial is nothing but a polynomial function with the odd numbers as powers. The general form of polynomial function is given as ax 2 + bx + c = 0. The odd polynomial in the sense the variables in the given functions have the power as odd terms. The general form of odd polynomial function is given as
ax 5 + b x 3 + cx + d = 0
The above function is named as odd polynomial function. The power terms present in the variable x are odd numbers. The following are the example problems for odd polynomials.
Example problems on odd polynomials:
Example 1:
Solve the odd polynomial function.
f(u) = u5 – 3u 3 – 4 u + u
Solution:
The given function is
f(u) = u5 – 3u 3 – 4 u + u
Differentiate the above odd polynomial function, we get
f '(u) = 5 u 4 – 3(3u 2 ) – 4( 1 ) + 1
f '(u) = 5u 4 – 9u 2 – 4 + 1
f '(u) = 5u 4 – 9u 2 – 3 is the solution.
Example 2:
Solve the odd polynomial function.
∫f(u) = u5+u3+18u + 12 du
Solution:
The given odd polynomial function is
∫f(u) = u5+u3+18u +12 du
∫f(u) = ∫ (u5+u3+18u +12) du
∫f(u) = ∫u5 du +∫u3 du+ ∫ 18u du +∫ 12
Integrate the above odd polynomial function, we get
F(u) =u6/6 + u4/4 + 18 u2 /2 + 12u
Simplifying the above function we get
F(u) =u6 / 6+ u4 / 4 +9u2 + 12u is the solution.
Example 3:
Solve the odd Polynomial Functions Examples.
∫f(u) = 4u3+3u+10 du
Solution:
The given odd polynomial function is
∫f(u) = 4u3+3u+10 du
∫f(u) = ∫ (4u3+3u+10) du
∫f(u) = ∫4u3 du +∫3u du+ ∫ 10
Integrate the above odd polynomial function, we get
F(u) =4u4/4 + 3u2/2 + 10 u
Simplifying the above function we get
F(u) =u4 + 3u2 /2+10u is the solution.
Example 4:
Solve the odd polynomial function.
∫f(u) = 4u3+3u+10u du
Solution:
The given odd polynomial function is
∫f(u) = 4u3+3u+10u du
∫f(u) = ∫ (4u3+3u+10u) du
∫f(u) = ∫4u3 du +∫3u du+ ∫ 10u du
Integrate the above odd polynomial function.
We get
F (u) =4u4/4 + 3u2/2 + 10 u2 /2
Simplifying the above function we get
F(u) =u4+ 3u2 / 2 +5u2
F(u) =u4+ 13u2 / 2 is the solution.
Practice problems on odd polynomials:
1) Solve the odd polynomial function.
f(x) = 4x 3 – 4 x + 5
Answer: f '(x) = 12x 2 + 4 is the solution.
2) Solve the odd polynomial function.
∫f(x) = x5+10x dx
Answer: F(x) =x6 / 6+ 5x2 is the solution.