Prove that there is a number that is exactly one more than its cube. (don’t solve just show there is one)
Prove that the function f(x)= cosx-x has a zero in (o. pi/2) Justify.
x = x^3 + 1
x^3 - x + 1 = 0
let f(x) = x^3 - x + 1
every cubic function, just like every odd exponent equation, crosses the x-axis at least once.
BTW, how about x = appr. -1.3247
for cosx - x = 0
cosx = x
graph y = cosx an y = x on the same graph
they only cross once, hence one solution
To prove that there is a number that is exactly one more than its cube, we need to find a number 'x' such that x = x^3 + 1. We can algebraically solve this equation to show the existence of such a number.
1. Start with the equation x = x^3 + 1.
2. Rearrange the equation to obtain x^3 - x + 1 = 0.
3. Now, we consider the function f(x) = x^3 - x + 1.
4. We want to show that there exists a value of 'x' for which f(x) = 0.
5. We can use various methods to prove this, such as the Intermediate Value Theorem or by showing that f(x) is continuous and changes sign within a certain interval.
To prove that the function f(x) = cos(x) - x has a zero in the interval (0, pi/2), we can use a similar approach:
1. Consider the function f(x) = cos(x) - x.
2. We want to show that there exists a value of 'x' in the interval (0, pi/2) for which f(x) = 0.
3. First, note that f(0) = cos(0) - 0 = 1 and f(pi/2) = cos(pi/2) - (pi/2) = 0 - (pi/2) = -pi/2 < 0.
4. Since f(x) is a continuous function, and f(0) and f(pi/2) have different signs, by the Intermediate Value Theorem, there exists a value of 'x' in the interval (0, pi/2) where f(x) = 0.
5. Therefore, we can conclude that the function f(x) = cos(x) - x has a zero in the interval (0, pi/2).