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First Derivative Test for Local Maximum

Let Px=a0+a1x...

Question

Let $P (x) = a_{0} + a_{1} x^{2} + a_{2} x^{4} + . . . . . . + a_{n} x^{2 n}$ be a polynomial in a variable x with $0 < a_{0} < a_{1} < a_{2} < a_{3} . . . . . . . < a_{n}$ . The function P(x) has

A

neither a maximum nor a minimum

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B

only one maximum

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C

only one minimum

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D

only one maximum and only one minimum

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E

none of these

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Solution

The correct option is C only one minimum
$We have, P (x) = a_{0} + a_{1} x^{2} + a_{2} x^{4} + . . . . . . + a_{n} x^{2 n} ∴ P' (x) = 2 a_{1} x + 4 a_{2} x^{3} + . . . . . . + 2 n a_{n} x^{2 n - 1} \Rightarrow P' (x) = x (2 a_{1} + 4 a_{2} x^{2} + . . . . . . + 2 n a_{n} x^{2 n - 2}) Clearly, P' (x) > 0 for x > 0 and P' (x) < 0 for x < 0 . Thus, P (x) increases for all x > 0 and decreases for all x < 0 . Also, P (x) cannnot be less than 0 as all the coefficients are positive and powers of variable x are all even . ∴ P' (x) has x = 0 as the point of minima .$

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Similar questions

P (x) = a_{0} + a_{1} x^{2} + a_{2} x^{4} + . . . + a_{n} x^{2 n}

be a polynomial in a real variable x with

0 < a_{0} < a_{1} < a_{2} < . . . < a_{n} .

The function P(x) has

P (x) = a_{0} + a_{1} x^{2} + a_{2} x^{4} + . . . + a_{n} x^{2 n}

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