State whether the equation is true, false, or an open sentence.

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

2 years ago

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What is the simplified form of the quantity of x plus 4, all over 5 + the quantity of x plus 5, all over 5?

Novay_Z [31]

Thats the right answer.

6 0

3 years ago

Carlotta subscribes to the HotBurn music service. She can download no more than 11 song files per week. Carlotta has already dow

IceJOKER [234]

That is how I solved that :)

8 0

3 years ago

Solve the following problem when -2 is outside an absolute value equation: -2|-2r-4|=-12

VMariaS [17]

-2|-2r-4|=-12

divide by -2

|-2r-4| = 6

we get a positive and negative solution

-2r-4 = 6 and -2r - 4 = -6

-add 4 to each side

-2r = 10 -2r = -2

divide by -2

r = -5 and r = 1

Answer r = -5 ,1

8 0

3 years ago

Read 2 more answers

Write the main formulas in perimeter and area in 7ty grade

nadya68 [22]

Answer:

Triangles and Quadrilaterals

P = a+b+c . . . . triangle with sides a, b, c
P = a+b+c+d . . . . general quadrilateral with side lengths a, b, c, d
P = 2(L+W) . . . . rectangle with length L and width W
A = 1/2bh . . . . triangle with base b and height h
A = (1/2)ab·sin(C) . . . . triangle given sides a, b and included angle C
A = √(s(s-a)(s-b)(s-c)) . . . . triangle with sides a, b, c, with s=P/2. "Heron's formula"
A = bh . . . . parallelogram with base b and height h. Includes rectangle and square.
A = ab·sin(θ) . . . . parallelogram with adjacent sides a, b, and included angle θ
A = 1/2(b1+b2)h . . . . trapezoid with parallel bases b1, b2 and height h
A = √((s-a)(s-b)(s-c)(s-d)) . . . . area of cyclic quadrilateral with sides a, b, c, d and s=P/2. "Brahmagupta's formula"

Circles

C = πd = 2πr . . . . circumference of a circle of radius r or diameter d
A = πr² . . . . area of a circle of radius r
A = (1/2)r²θ . . . . area of a circular sector with radius r and central angle θ radians
A = 1/2rs . . . . area of a circular sector with radius r and arc length s

Solids (3-dimensional objects)

A = 2(LW +LH +WH) = 2(LW +H(L+W)) . . . . surface area of a rectangular prism of length L, width W, height H
A = 4πr² . . . . area of a sphere of radius r

Step-by-step explanation:

Perimeter

The perimeter is the sum of the lengths of the sides of a plane figure. When the sides are the same length, multiplication can take the place of repeated addition. Of course, opposite sides of a parallelogram (includes rhombus, rectangle, and square) are the same length, as are adjacent sides of a rhombus (includes square).

The perimeter of a circle is called it circumference. The ratio of the circumference to the diameter is the same for all circles, a transcendental constant named pi (not "pie"), the 16th letter of the Greek alphabet (π). The value of pi is sometimes approximated by 22/7, 3.14, 3.1416, or 355/113. The last fraction is good to 6 decimal places. It has been calculated to several trillion digits.

Area

Fundamentally, the area of a figure is the product of two orthogonal linear dimensions. For odd-shaped figures, the area can be decomposed into smaller pieces that can be added up. (Calculus is used to do this for areas with irregular boundaries.)

The most common figures for which we find areas are triangles, rectangles, and circles. We sometimes need the area of a fraction of a circle, as when we're computing the lateral area of a cone.

It may help you remember the formulas if you notice the similarity of formulas for area of a triangle and area of a circular sector.

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Among the formulas above are some that give area when sides and angles are known. The special case of a cyclic quadrilateral (one that can be inscribed in a circle) has its own formula. The similar formula for the area of a triangle from side lengths can be considered to be a special case of the quadrilateral formula where one side is zero.

The formula for area of a trapezoid is somewhat interesting. If the two bases are the same length, the figure is a parallelogram, and the formula matches that for a parallelogram. If one of the bases is zero length, the figure is a triangle, and the formula matches that of a triangle.

In any event, it is useful to note that the area is the product of height and average width. This will be true of any figure — a fact that is used to find the average width in some cases.

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Comment on maximum area, minimum perimeter

A polygon will have the largest possible area for a given perimeter, or the smallest possible perimeter for a given area, if it is a regular polygon. For a quadrilateral, the largest area for a given perimeter is that of a square. For a given perimeter, a regular polygon with more sides will have a larger area.

As the number of sides increases toward infinity, the polygon increasingly resembles a circle, which has the largest possible area for a given perimeter.

3 0

3 years ago