Let $$f: [0, 1] \to \mathbb{R}$$ 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 $$x \in (0, 1)$$, at which $$f''(x) = 0$$, is

We have $$f: [0,1] \to \mathbb{R}$$, twice differentiable in $$(0,1)$$, with $$f(0) = 3$$ and $$f(1) = 5$$. The line $$y = 2x + 3$$ intersects the graph of $$f$$ at exactly two distinct points in $$(0,1)$$.

At $$x = 0$$: $$y = 3 = f(0)$$. At $$x = 1$$: $$y = 5 = f(1)$$.

So the line connects the endpoints of $$f$$.

Then $$h(0) = 0$$ and $$h(1) = 0$$.

The line intersects $$f$$ at exactly two points in $$(0,1)$$, so $$h(x) = 0$$ at exactly two points in $$(0,1)$$. Combined with the endpoints, $$h$$ has at least 4 zeros in $$[0,1]$$: $$x = 0$$, two interior points, and $$x = 1$$.

With 4 zeros of $$h$$, by Rolle's theorem, $$h'(x) = 0$$ at least 3 times in $$(0,1)$$.

Applying Rolle's theorem again to $$h'$$, we get $$h''(x) = 0$$ at least 2 times in $$(0,1)$$.

Since $$h''(x) = f''(x) - 0 = f''(x)$$, we conclude that $$f''(x) = 0$$ at least 2 times.

The least number of points where $$f''(x) = 0$$ is 2.

The answer is 2.