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soldi70 [24.7K]
3 years ago
12

For the function F(x)=1/x+1, whose graph is shown below, what is the relative value of F(x) when the value of x is close to -1.

Mathematics
1 answer:
daser333 [38]3 years ago
6 0
Well as x can never actually be -1 because I'm in the denominator -1 + 1 = 0 and we cannot divide by zero. But we can look at what number it approaches and i assume that is the relative value. sometimes functions will have asymptotes and others will have holes in the graph. this one would have an asymptote going down at a rapid rate. the asymptote would go on forever getting infinitely close to -1 but never touching. So I would say since the asymptote goes down forever that the graph approaches negative infinity
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What is 5✓28 +✓63 In simplest radical form<br>I need it asap.
Ksenya-84 [330]

Answer:

13✓7

Step-by-step explanation:

5✓28 + ✓63

Break the radical into perfect squares.


5 times ✓4 times ✓7  +  ✓9 times ✓7

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3 years ago
Find f(4) please, I need help
mixer [17]

Answer:

f(4) = 1

Step-by-step explanation:

f(4) = what?

So if we know that x, aka 4, is technicially f is greater then 2, so you can plug in x as 4. So that means 5(4) - 19 = 20 - 19 = 1, so f(4) = 1

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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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