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प्रश्न
Find intervals of concavity and points of inflection for the following functions:
f(x) = `1/2 ("e"^x - "e"^-x)`
उत्तर
f'(x) = `1/2 ("e"^x + "e"^-x)`
f''(x) = `1/2 ("e"^x - "e"^-x)`
f”(x) = 0
⇒ `1/2 ("e"^x - "e"^-x)` = 0
Critical point x = 0
The intervals are `(-oo, 0)` and (`0, oo)`
In the interval `(-oo, 0)`, f”(x) < 0 ⇒ curve is concave down.
In the interval `(0, oo)`, f(x) > 0 ⇒ curve is concave up.
∴ The curve is concave up in `(0, oo)` and concave down in `(-oo, 0)`.
f'(x) changes its sign when passing through x = 0
Now f(0) = – (e° – e°)
= `1/2 (1 - 1)`
= 0
∴ The point of inflection is (0, 0).
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