1 Answer | Add Yours
The intermediate value theorem states that if a function f(x) is continuous in the interval [a, b] there exists a value of x for which f(x) = c where c lies in [a, b]
Take x = -1, h(-1) = 2*(-1) - 5 - 1 = -8
For x = 1, h(1) = 2 + 5 - 1 = 6
As 0 lies in the interval [-1, 1], there is a value of x for which f(x) = 0.
Rolle's theorem states that if a differentiable function has equal values at two different points, there is one point between the two where the first derivative of the function is equal to 0.
h(x) = 2x^3 + 5x - 1
h'(x) = 6x^2 + 5
The value of h'(x) = 6x^2 + 5 is greater than 5. It is never equal to 0. This implies that the function h(x) = 2x^3 + 5x - 1 is equal to 0 at only one point. The function has only one root.
The root of h(x) = 2x^3 + 5x - 1 lies in [-1, 1] and the function has only one real root.
We’ve answered 317,783 questions. We can answer yours, too.Ask a question