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Question
Prove that the following function do not have maxima or minima:
h(x) = x3 + x2 + x + 1
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Solution
Given function, h(x) = x3 + x2 + x + 1
∴ h'(x) = 3x2 + 2x + 1 x ∈ R.
if, 3x2 + 2x + 1 = 0 then,
x `= (-2 pm sqrt(4 - 12))/6`
x `= (-2 pm sqrt(-8))/6`
`= (-1 pm sqrt (-2))/3` ...[which is non-real]
For x ∈ R, h'(x) ≠ 0
Hence there is no highest or lowest value of h.
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