Intermediate Value Theorem Problem: Numbers equal to twice their cube?

Question

The number zero has the property that it is equal to twice its cube (0 = 2(0)^3). Use the Intermediate Value Theorem to show that there's another number with this property. Use algebra to find all numbers with this property.

Answer 1

Let x = the number
x = 2x^3
Subtract x from each side
0 = 2x^3 -x

Let f(x) = 2x^3 -x

Let x = -1, then
f(-1) = 2(-1)^3 -(-1)
= -2 + 1
= -1

Let x = -1/2, then
f(-1/2) = 2(-1/2)^3 - (-1/2)
= 2(-1/8) + 1/2
= -1/4 + 2/4
= 1/4

Since f(-1) < 0 < f(-1/2), then there must be a value in between where f(x) = 0

To find the solutions, you can factor.
2x^3 -x = 0
x(2x^2 -1) = 0
x = 0 or 2x^2 -1 = 0
x = 0, x = -√(1/2), and x = √(1/2)

Answer 2

Consider the function f(x) = 2x^3 - x. Then f(1/2) < 0 and f(5/6) > 0 so there is a number between 1/2 and 5/6 such that f(x) = 0.

Now, algebraically, we have f(x) = 0 implies
2x^3 - x = 0 implies x(2x^2 - 1) = 0 implies x = 0 or
2x^2-1 = 0 implies x = (plus or minus) 1/sqrt(2).

Answer 3

f(47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0) = -5,-2(47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0)^3 +-5,-2(47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0)^47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c04(47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0)+6 = 47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0 +47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0 +47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0 + 6 =347827d4dac24cdd7a622544b01834c06 f(47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0) = -5,-2(47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0)^3 +-5,-2(47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0)^47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c04(47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0)+6 = 47827d4dac24cdd7a622544b01834c06+8+8+6 =38 The era (347827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0347827d4dac24cdd7a622544b01834c06) does not incorporate 47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0 Therefore47827d4dac24cdd7a622544b01834c0 the function does not have a nil on [47827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c047827d4dac24cdd7a622544b01834c0] in accordance to intermediate cost theorem47827d4dac24cdd7a622544b01834c0

Answer 4

Have never heard of that theorem.

but x=2(x)^3 or 2(x)^3-x=0
x*(2x^2-1)=0
x=0 or x^2=1/2 , x=+-sqrt(2)/2

Answer 5

+/- sqrt(2)/2 satisfy f(x) = 2x^3-x = 0.
0 is the only other real number that satisfies this function.
f(x) is continuous over the real numbers.
Henece the intermediate value theorem applies.

Answer 6

algebra part is N = 2N^3, which means
2N^3 - N = 0
solve for N we get: N (2N^2-1) = 0

which yields N=0 for first part and,
2N^2-1 = 0 or N = sqrt2 / 2 = .707 for the 2nd part

Answer 7

x = 2x^3
2x^3 - x = 0
x*(2x^2 -1) = 0
x = 0, +/- sqrt(2) / 2