# The Intermediate Value Theorem

The Intermediate Value Theorem ensures that for a continuous function, any value between its outputs at two points is also achieved somewhere within that interval. It confirms the existence of solutions without pinpointing their exact location.

When the function is continuous, it means there are no jumps, breaks, or gaps in its graph. In this context, the Intermediate Value Theorem can be applied. The continuity ensures that for any value between the function’s outputs at two points in a given interval, there is at least one corresponding input within that interval where the function takes that value. This continuity is crucial for the Intermediate Value Theorem as it guarantees that the function smoothly transitions through all values between its outputs at the interval’s endpoints, thus making the theorem applicable.

to use this theorem, we can follow these steps:

To use the Intermediate Value Theorem (IVT), you can follow these basic steps:

1. Verify Continuity: Ensure the function $$f(x)$$ is continuous on the closed interval $$[a, b]$$ .
2. Identify Endpoints: Evaluate the function at the endpoints of the interval, $$f(a)$$ and $$f(b)$$.
3. Check Intermediate Value: Determine if the value you’re looking for, say $$d$$ , lies between $$f(a)$$ and ( f(b) ).
4. Conclude Existence: If $$d$$ is between $$f(a)$$ and $$f(b)$$, then IVT guarantees that there is at least one $$c$$ in $$[a, b]$$ such that $$f(c) = d$$.

Here is an example:

Let’s consider the function $$f(x) = x^3 – x$$ on the interval $$[-2, 2]$$.

1. Check Continuity:
• $$f(x) = x^3 – x$$ is a polynomial, and polynomials are continuous everywhere. Therefore, $$f(x)$$ is continuous on $$[-2, 2]$$.
1. Choose a Value $$d$$ Between $$f(a)$$ and $$f(b)$$:
• Let’s calculate $$f(-2)$$ and $$f(2)$$:
• $$f(-2) = (-2)^3 – (-2) = -8 + 2 = -6$$
• ( f(2) = 2^3 – 2 = 8 – 2 = 6 )
• Choose $$d = 0$$, which clearly lies between $$-6$$ and $$6$$.
1. Apply IVT:
• Since $$0$$ is between $$f(-2) = -6$$ and $$f(2) = 6$$, and $$f(x)$$ is continuous on $$[-2, 2]$$, the IVT guarantees there exists at least one $$c$$ in $$[-2, 2]$$ such that $$f(c) = 0$$.
1. Finding $$c$$:
• To find the actual value of $$c$$, solve $$x^3 – x = 0$$.
• Factoring out $$x$$, we get $$x(x^2 – 1) = 0$$, so the roots are $$x = 0, 1, -1$$.
• All these roots are within the interval $$[-2, 2]$$, satisfying the theorem.

In this example, the IVT tells us that the function $$f(x) = x^3 – x$$ takes on every value between $$-6$$ and $$6$$ at least once in the interval $$[-2, 2]$$, including the value $$0$$.

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