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Let f(x) be a real-valued function such that f(x0) = 0 for some x
0 
∈ (0,1) and f(x) > 0 for all x∈(0,1). Then f(x) has
  • a)
    two distinct local minima in (0,1)
  • b)
    exactly one local minimum in (0,1)
     
  • c)
    one local maximum in (0,1)
  • d)
    no local minimum in (0,1)
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Let f(x) be a real-valued function such that f(x0) = 0 for some x0&isi...
x0 ∈ (0,1), where f(x) = 0 is stationary piont
and f(x) > 0                    
So                                    f(x0) = 0
and                                  f(0) > 0, where x0 ∈(0,1)
Hence, f(x) has exactly one local minima in (0,1)
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Community Answer
Let f(x) be a real-valued function such that f(x0) = 0 for some x0&isi...
If f(x0) = 0, then x0 is a root of the function f(x). This means that when x = x0, the value of f(x) is 0.

In other words, f(x0) = 0 can be written as f(x) = 0 when x = x0.

This information tells us that the function f(x) has at least one root at x = x0. It does not provide any information about other possible roots or the behavior of the function at other points.

To learn more about the function f(x), additional information is needed, such as the properties of the function, its derivative, or other conditions.
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Let f(x) be a real-valued function such that f(x0) = 0 for some x0∈ (0,1) and f(x) > 0 for all x∈(0,1). Then f(x) hasa)two distinct local minima in (0,1)b)exactly one local minimum in (0,1)c)one local maximum in (0,1)d)no local minimum in (0,1)Correct answer is option 'B'. Can you explain this answer?
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