If a continuous function f(x) does not have a root in the interval [a, b], then which one of the following statements is TRUE?

Understanding the Continuous Function Property

The question asks us to identify a true statement about a continuous function, let's call it $f(x)$, over a specific interval $[a,b]$. The key piece of information is that this function does not possess a root within this interval. A root of a function is defined as a value $x_0$ where the function equals zero, meaning $f(x_0)=0$. The interval $[a,b]$ includes both endpoints, $a$ and $b$. We need to figure out the relationship between the function's values at these endpoints, $f(a)$ and $f(b)$, given the absence of a root.

Key Mathematical Concepts

Continuous Function

A function $f(x)$ is described as continuous on an interval if its graph can be drawn without lifting your pen from the paper. In more technical terms, for any point within the interval, the function's value at that point must be equal to the limit of the function as it approaches that point. This property ensures smooth, unbroken behavior.

Root of a Function

A root, also known as a zero, of a function $f(x)$ is an input value $x_0$ for which the output $f(x_0)$ is equal to 0. Graphically, roots represent the points where the function's graph crosses or touches the x-axis.

The Intermediate Value Theorem (IVT)

This problem relies heavily on the Intermediate Value Theorem (IVT). The IVT states that if a function $f(x)$ is continuous on a closed interval $[a,b]$, and we consider a value $k$ that lies between $f(a)$ and $f(b)$, then the function must take on this value $k$ at some point $c$ within the interval $[a,b]$. That is, $f(c)=k$ for some $c\in [a,b]$.

A very important consequence of the IVT relates directly to finding roots: If $f(x)$ is continuous on $[a,b]$ and the values $f(a)$ and $f(b)$ have opposite signs, this implies that $f(a)\cdot f(b)<0$. In this specific scenario, the IVT guarantees that there must be at least one root $c$ within the open interval $(a,b)$ where $f(c)=0$. This means the function must cross the x-axis between $a$ and $b$.

**Analyzing the Condition: No Root in $[a,b]$ **

We are given that the function $f(x)$ is continuous on $[a,b]$ and importantly, it has no root in this interval. This means that for all $x$ such that $a\le x\le b$, we have $f(x)\ne 0$. Let's apply the logic from the IVT:

- The IVT tells us that if $f(a)\cdot f(b)<0$, a root is guaranteed to exist within the interval $(a,b)$.
- However, we know that no such root exists in $[a,b]$.
- Therefore, the condition $f(a)\cdot f(b)<0$ cannot be true, because if it were true, it would contradict the given information that there is no root.

**Evaluating the Product $f(a)\cdot f(b)$ **

Since we've established that $f(a)\cdot f(b)<0$ must be false, we need to consider the other possibilities for the product $f(a)\cdot f(b)$:

- **Possibility 1: $f(a)\cdot f(b)=0$ ** This situation occurs if $f(a)=0$ or $f(b)=0$ (or both). If $f(a)=0$, it means $a$ is a root. If $f(b)=0$, it means $b$ is a root. The problem statement explicitly states that there is no root in the interval $[a,b]$. This explicitly excludes the endpoints $a$ and $b$. Thus, $f(a)$ cannot be 0, and $f(b)$ cannot be 0. This eliminates the possibility that $f(a)\cdot f(b)=0$.
- **Possibility 2: $f(a)\cdot f(b)>0$ ** This occurs when $f(a)$ and $f(b)$ have the same sign. Both could be positive, or both could be negative. When $f(a)$ and $f(b)$ share the same sign, the IVT does not guarantee that the function crosses the x-axis between $a$ and $b$. It's possible for the function to start and end above the x-axis (both $f(a)>0,f(b)>0$) and stay above it throughout the interval, or start and end below the x-axis (both $f(a)<0,f(b)<0$) and stay below it. This scenario perfectly aligns with the condition that there is no root in the interval $[a,b]$.

Given that $f(a)\cdot f(b)<0$ is false and $f(a)\cdot f(b)=0$ contradicts the problem conditions, the only logical conclusion is that $f(a)\cdot f(b)>0$ must be true.

Analysis of the Options

Let's review the given options in light of our analysis:

- **Option 1: $f\left(a\right).f\left(b\right)=0$ ** This is incorrect. As explained, this condition implies $f(a)=0$ or $f(b)=0$, meaning $a$ or $b$ is a root, contradicting the premise.
- **Option 2: $f\left(a\right).f\left(b\right)<0$ ** This is incorrect. The IVT states that this condition (along with continuity) guarantees a root exists in $(a,b)$, which contradicts the problem statement.
- **Option 3: $f\left(a\right).f\left(b\right)>0$ ** This is the correct statement. It signifies that $f(a)$ and $f(b)$ have the same sign, which is consistent with a continuous function not crossing the x-axis (having no root) within the interval $[a,b]$.
- **Option 4: $f\left(a\right)/f\left(b\right)\le 0$ ** This is incorrect. This inequality implies either $f(a)$ and $f(b)$ have opposite signs ($f(a)/f(b)<0$, equivalent to $f(a)\cdot f(b)<0$) or $f(a)=0$ (assuming $f(b)\ne 0$). Both cases imply the existence of a root, contradicting the problem statement. The case where $f(b)=0$ would also mean $b$ is a root.
- Option 5: (No statement provided) This option is not applicable.

Conclusion

To summarize, for a continuous function $f(x)$ that has no root in the interval $[a,b]$, the values $f(a)$ and $f(b)$ cannot be zero and must have the same sign. Therefore, the only statement that must be true is $f(a)\cdot f(b)>0$.