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 ___________.
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 ___________.
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 ___________.
Updated On: Oct 20, 2024
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Solution and Explanation
f' (a) = f' (b) = f' (c) = 2
⇒ f" (x) is zero
for atleast x1 ∈ (a, b) & x2 є (b, c)
⇒ f" (x) is zero
for atleast x1 ∈ (a, b) & x2 є (b, c)
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.