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Question
Choose the correct option from the given alternatives :
If f(x) = `(x^2 - 1)/(x^2 + 1)`, for every real x, then the minimum value of f is ______.
Options
1
0
–1
2
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Solution
If f(x) = `(x^2 - 1)/(x^2 + 1)`, for every real x, then the minimum value of f is –1.
Explanation:
f(x) = `(x^2 - 1)/(x^2 + 1) = (x^2 + 1 - 2)/(x^2 + 1) = 1 - 2/(x^2 + 1)`
Therefore, f(x) < 1 ∀x and ≥ –1 ......`(∵ 2/(x^2 + 1) ≤ 2)`
Therefore, –1 ≤ f(x) < 1
Hence, f(x) has minimum value −1 and also there is no maximum value.
Alter: We have
f'(x) = `((x^2 + 1)2x - (x^2 - 1)2x)/(x^2 + 1)^2 = (4x)/(x^2 + 1)^2`
f'(x) = 0
⇒ x = 0
Now, f"(x) = `((x^2 + 1)^2 4 - 4x.2(x^2 + 1)2x)/(x^2 + 1)^4`
= `((x^2 + 1)4 - 16x(x))/(x^2 + 1)^3`
= `(-12x^2 + 4)/(x^2 + 1)^3`
Therefore, f′′(0) > 0 and there is only one critical point that has minima. Hence, f(x) has the least value at x = 0
fmin = f(0) = –1/1 = –1
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