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chubhunter [2.5K]
4 years ago
13

Use Descartes' Rule of Signs to find the number of possible positive real roots and the number of possible negative real roots f

or the function f(x) = x^4+ 2x^3-3x^2- 8x - 4.
a positive 1; negative 3 or 1


b. positive 1; negative 3 or 5


C. positive 3; negative 3 or 1


d. positive 3; negative 3 or 5
Mathematics
1 answer:
Drupady [299]4 years ago
6 0

Answer:

a positive 1; negative 3 or 1

Step-by-step explanation:

To determine the number of positive roots, we have to determine the number of sign changes for f(x) = x⁴ + 2x³ - 3x² - 8x - 4.

The coefficients in f(x) are +1, +2, -3, -8, -4.

Since there is only one sign change from +2 to -3, we have 1 positive root.

To determine the number of negative roots, we have to determine the number of sign changes for f(-x) = (-x)⁴ + 2(-x)³ - 3(-x)² - 8(-x) - 4 = x⁴ - 2x³ - 3x² + 8x - 4

The coefficients in f(-x) are +1, -2, -3, +8, -4.

Since there is three sign change from +1 to -2, from -3 to +8, and from +8 to -4. So,we have 3 or 1 negative root, since the number of negative roots is equal to the number of sign changes or an even number less than the number of sign changes. So, 3 -2 = 1

So, the number roots are of positive 1; negative 3 or 1

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3:4 and 4:5 equivalent?
Sergeeva-Olga [200]

Step-by-step explanation:

As a short answer NO

As a complete answer there's a decimal diference of 0.3333333333

8 0
3 years ago
. Use Lagrange multipliers to find the maximum and minimum values of the function, f, subject to the given constraint, g. (Place
zzz [600]

Answer:

Minimum value of f(x, y, z) = (1/3)

Step-by-step explanation:

f(x, y, z) = x⁴ + y⁴ + z⁴

We're to maximize and minimize this function subject to the constraint that

g(x, y, z) = x² + y² + z² = 1

The constraint can be rewritten as

x² + y² + z² - 1 = 0

Using Lagrange multiplier, we then write the equation in Lagrange form

Lagrange function = Function - λ(constraint)

where λ = Lagrange factor, which can be a function of x, y and z

L(x,y,z) = x⁴ + y⁴ + z⁴ - λ(x² + y² + z² - 1)

We then take the partial derivatives of the Lagrange function with respect to x, y, z and λ. Because these are turning points, each of the partial derivatives is equal to 0.

(∂L/∂x) = 4x³ - λx = 0

λ = 4x² (eqn 1)

(∂L/∂y) = 4y³ - λy = 0

λ = 4y² (eqn 2)

(∂L/∂z) = 4z³ - λz = 0

λ = 4z² (eqn 3)

(∂L/∂λ) = x² + y² + z² - 1 = 0 (eqn 4)

We can then equate the values of λ from the first 3 partial derivatives and solve for the values of x, y and z

4x² = 4y²

4x² - 4y² = 0

(2x - 2y)(2x + 2y) = 0

x = y or x = -y

Also,

4x² = 4z²

4x² - 4z² = 0

(2x - 2z) (2x + 2z) = 0

x = z or x = -z

when x = y, x = z

when x = -y, x = -z

Hence, at the point where the box has maximum and minimal area,

x = y = z

And

x = -y = -z

Putting these into the constraint equation or the solution of the fourth partial derivative,

x² + y² + z² = 1

x = y = z

x² + x² + x² = 1

3x² = 1

x = √(1/3)

x = y = z = √(1/3)

when x = -y = -z

x² + y² + z² = 1

x² + x² + x² = 1

3x² = 1

x = √(1/3)

y = z = -√(1/3)

Inserting these into the function f(x,y,z)

f(x, y, z) = x⁴ + y⁴ + z⁴

We know that the two types of answers for x, y and z both resulting the same quantity

√(1/3)

f(x, y, z) = x⁴ + y⁴ + z⁴

f(x, y, z) = (√(1/3)⁴ + (√(1/3)⁴ + (√(1/3)⁴

f(x, y, z) = 3 × (1/9) = (1/3).

We know this point is a minimum point because when the values of x, y and z at turning points are inserted into the second derivatives, all the answers are positive! Indicating that this points obtained are

S = (1/3)

Hope this Helps!!!

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Step-by-step explanation:

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