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प्रश्न
Show that the local maximum value of `x + 1/x` is less than local minimum value.
उत्तर
Let y = `x + 1/x`
⇒ `"dy"/"dx" = 1 - 1/x^2`
`"dy"/"dx"` = 0
⇒ x2 = 1
⇒ x = ± 1.
`("d"^2y)/("dx"^2) = + 2/x^3`
Therefore `("d"^2y)/("dx"^2)` (at x = 1) > 0 and `("d"^2y)/("dx"^2)` (at x = –1) < 0.
Hence local maximum value of y is at x = –1 and the local maximum value = – 2.
Local minimum value of y is at x = 1 and local minimum value = 2.
Therefore, local maximum value (–2) is less than local minimum value 2.
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