Question

(b) show that f(0)<0<f(5) and use the intermediate - value theorem to conclude that there exists a number c∈(0,5) such that f(c)=0. since f(0)=□ and f(5)=□, this implies that f(0)<0<f(5).

Explanation:

Step1: Evaluate f(0)

We need the function f(x) to find f(0). Since it's not given, assume f(x)=x - 1 for illustration. Then f(0)=0 - 1=-1.

Step2: Evaluate f(5)

Using the assumed f(x)=x - 1, f(5)=5 - 1 = 4.

Step3: Apply Intermediate - Value Theorem

The Intermediate - Value Theorem states that if a function 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 = 0, b = 5, k = 0, f(0)=-1<0 and f(5)=4>0. So there exists c∈(0,5) such that f(c)=0.

Answer:

Since we assumed f(x)=x - 1, f(0)=-1 and f(5)=4. (In a real - problem, you need to use the actual given function to calculate f(0) and f(5)).