Intermediate Value Theorem?

Question

Use the Intermediate Value Theorem to show that the polynomial function below has a zero in the given interval.

f(x) = x^4 + 8x^3 - x^2 + 2; [-1, 0 ]

Answer 1

My understanding of the Intermediate Value Thm is that if you put those two values in, -1 and 0, and the function which = y, is positive for one and negative for the other, then it had to have crossed the x-axis. And, the intersection with the x-axis is your zero.

f(-1) = 1 -8 -1 + 2 = -6

f(0) = + 2

So, yes, there must be a zero in that interval.

that's it! :)

Answer 2

The intermediate theorum basically defines a continuous function. Let's look at something simple in two dimensions. Pick some limits on the x axis. They usually use a and b, let's use 3 and 7. Now draw a curve which is continuous on the interval between 3 and 7. A straight line will work or any curve. Make sure the curve (line) has no breaks and doesn't do anything nuts like go off the graph to infinity in the interval from 3 to 7. This called a continuous function -- it CONTINUES through the interval with no breaks and the function is defined by some equation.. The intermediate value theorum says that if you pick a number between a and b (3 and 7) and plug it into the equation, the result will fall on the curve. It's as simple as that.

Answer 3

The Intermediate Value Theorem says that a continuous
function on an interval [a,b] takes all the values between f(a)
and f[b] (Note the function might take more values, the theorem
says that at least it takes all the values between f(a) and f(b).)

Here f(0) is positive and and f(-1) is negative, so there
is (at least) a c in the open interval (-1,0) so that f(c)=0.

Answer 4

f(-1) = 1 - 8 - 1 + 2 = -6
f(0) = 0 + 0 - 0 + 2 = +2

The function is continuous over the interval (thre are no singularities, no jumps, no discontinuities).
Therefore, there must be a value "r" in the interval such that f(r) = 0

Answer 5

f(x) is continuous on the interval [-1, 0] and differentiable on the interval (-1, 0).

f(-1) = (-1)^4 + 8(-1)^3 - (-1)^2 + 2 = 1 - 8 - 1 + 2 = -6 < 0
f(0) = (0)^4 + 8(0)^3 - (0)^2 + 2 = 2 > 0

f(-1) < 0 (below the x-axis)
f(0) > 0 (above the x-axis)

The function is continuous on [-1, 0] so it has to pass through the x-axis where f(x) = 0. So, it has at least one root in [-1,0]