Let f : [0, 1] → R be a twice differentiable function in (0, 1) such that f(0) = 3 and f(1) = 5. If the line y = 2x + 3 intersects the graph of f at only two distinct points in (0, 1), then the least number of points x (0, 1), at which f"(x) = 0, is_____.Read more on Sarthaks.com - https://www.sarthaks.com/3214806/let-f-0-1-r-be-a-twice-differentiable-function-in-0-1-such-that-f-0-3-and

Question

Let f : [0, 1] → R be a twice differentiable function in (0, 1) such that f(0) = 3 and f(1) = 5. If the line y = 2x + 3 intersects the graph of f at only two distinct points in (0, 1), then the least number of points x (0, 1), at which f"(x) = 0, is_____.Read more on Sarthaks.com - https://www.sarthaks.com/3214806/let-f-0-1-r-be-a-twice-differentiable-function-in-0-1-such-that-f-0-3-and

Answer 1

Given f(x) cuts y = 2x + 3 at two distinct point between x ∈ (0, 1).

So, f'(a) = f'(b) = f'(c) = 2 {as slope of given line y = 2x + 3 is 2}

So we can conclude that f''(x) is zero for atleast x₁ ∈ (a, b) & x₂ ∈ (b, c).