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 = 2 x + 3 intersects the graph of f at only two distinct points in 0 , 1 , then the least number of points x ∈ 0 , 1 , a

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Let f : 0 , 1 &#8594; R be a twice differentiable function in 0 , 1 such that f 0 = 3 and f 1 = 5 . If the line y = 2 x + 3 intersects the graph of f at only two distinct points in 0 , 1 , then the least number of points x &#8712; 0 , 1 , at which f ' ' x = 0 , is

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Correct answer: 2

Let f : 0 , 1 &#8594; R be a twice differentiable function in 0 , 1 such that f 0 = 3 and f 1 = 5 . If the line y = 2 x + 3 intersects the graph of f at only two distinct points in 0 , 1 , then the least number of points x &#8712; 0 , 1 , at which f ' ' x = 0 , is Now plotting the diagram of given data we have, Given f x cuts y = 2 x + 3 at two distinct point between x &#8712; 0 , 1 , So, f ' a = f ' b = f ' c = 2 as slope of given line y = 2 x + 3 is 2 So we can conclude that f ' ' x is

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 = 2 x + 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

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Let f : 0 , 1 &#8594; R be a twice differentiable function in 0 , 1 such that f 0 = 3 and f 1 = 5 . If the line y = 2 x + 3 intersects the graph of f at only two distinct points in 0 , 1 , then the least number of points x &#8712; 0 , 1 , at which f &#39; &#39; x = 0 , is

Correct answer

Step-by-step solution

More from Application of Derivatives

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 = 2 x + 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