Is there any real number exactly one less than its cube intermediate value theorem?

≈ 1.32471795725...

Explanation:

If x is one less than its cube, then

x = x³ - 1,

x³ - x - 1 = 0

so f(x) = ? would be an appropriate (continuous) function to apply the Intermedate Value Theorem on some appropriate interval to see if it takes on the value 0 in that interval.

f(x) = x³ - x - 1

For large x the left hand side is positive, for x = 0 it is negative. The root can be calculated exactly, it is given by:

≈ 1.32471795725...

answer

explanation

the area of a circle is calculated using the formula

or

we can use any of the above formulae to calculate the required. preferably, let us use the latter since the diameter is

we substitute the diameter into the formula to get:

this is the exact value for the area of the circle.

the correct answer is a

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3