Question

a continuous function y = f(x) is known to be negative at x = 2 and positive at x = 8. why does the equation f(x)=0 have at least one solution between x = 2 and x = 8? illustrate with a sketch.

Explanation:

Step1: Recall Intermediate - Value Theorem

The Intermediate - Value Theorem states that if \(y = f(x)\) is continuous on a closed interval \([a,b]\), and \(k\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\) in the interval \((a,b)\) such that \(f(c)=k\). Here, \(a = 2\), \(b = 8\), \(f(2)<0\), \(f(8)>0\), and \(k = 0\).

Step2: Analyze the sketches

We need a sketch where the function \(y = f(x)\) is negative at \(x = 2\) (below the \(x\) - axis) and positive at \(x = 8\) (above the \(x\) - axis). Since the function is continuous, it must cross the \(x\) - axis (where \(y = 0\)) at least once between \(x = 2\) and \(x = 8\).

Answer:

The correct sketch is one where the curve representing \(y = f(x)\) starts below the \(x\) - axis at \(x = 2\) and ends above the \(x\) - axis at \(x = 8\) and is a continuous curve. Without seeing the actual content of options A, B, C, D in detail, the correct option should show a continuous curve that goes from below the \(x\) - axis at \(x = 2\) to above the \(x\) - axis at \(x = 8\) and crosses the \(x\) - axis at least once in the interval \((2,8)\).