Prove or Disprove:Let f : [0, 1] → R be continuous and satisfies f(0) &gt; 0 and f(1) &gt; 0. Then there is no c ∈ (0, 1) such that f(c) = 0

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Prove or Disprove:Let f : [0, 1] → R be continuous and satisfies f(0) &amp;gt; 0 and f(1) &amp;gt; 0. Then there is no c ∈ (0, 1) such that f(c) = 0

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To prove or disprove the statement, let&#039;s assume there exists a function f : [0, 1] → R that is continuous, satisfies f(0) &amp;gt; 0 and f(1) &amp;gt; 0, and there exists a c ∈ (0, 1) such that f(c) = 0.If f(c) = 0, it means that the function f crosses the x-axis at the point (c, 0). Since f(0) &amp;gt; 0 and f(1) &amp;gt; 0, the function f is positive on the interval [0, 1].According to the Intermediate Value Theorem, if a function f is continuous on a closed interval [a, b] and k is any number between f(a) and f(b), then there exists at least one c in the interval (a, b) such that f(c) = k.In our case, since f(0) &amp;gt; 0 and f(1) &amp;gt; 0, and f(c) = 0, we have a contradiction. The Intermediate Value Theorem states that if the function is continuous and positive at the endpoints, it cannot cross the x-axis and become zero within the interval.Therefore, the statement is true. There is no c ∈ (0, 1) such that f(c) = 0 for a function f : [0, 1] → R that is continuous and satisfies f(0) &amp;gt; 0 and f(1) &amp;gt; 0.

Prove or Disprove:Let f : [0, 1] →

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