If P(x) = a_0 + a_1x^2 + a_2x^4 + dots + a_nx | vedclass

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(B) Given $P(x) = a_0 + a_1x^2 + a_2x^4 + \dots + a_nx^{2n}$.For maxima and minima,we find the derivative $P'(x) = 0$.$P'(x) = 2a_1x + 4a_2x^3 + \dots

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Exactly one minimum point

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(B) Given $P(x) = a_0 + a_1x^2 + a_2x^4 + \dots + a_nx^{2n}$.For maxima and minima,we find the derivative $P'(x) = 0$.$P'(x) = 2a_1x + 4a_2x^3 + \dots + 2na_nx^{2n-1} = 2x(a_1 + 2a_2x^2 + \dots + na_nx^{2n-2}) = 0$.This gives $x = 0$ as a critical point.Now,we find the second derivative $P''(x) = 2a_1 + 12a_2x^2 + \dots + 2n(2n-1)a_nx^{2n-2}$.At $x = 0$,$P''(0) = 2a_1$.Since $0  0$.Since the second derivative is positive at $x = 0$,$P(x)$ has a local minimum at $x = 0$.Since $P(x)$ is an even function and the coefficients are strictly increasing,$x = 0$ is the only point where $P'(x) = 0$,thus $P(x)$ has exactly one minimum point.

If $P(x) = a_0 + a_1x^2 + a_2x^4 + \dots + a_nx^{2n}$ is a polynomial in $x \in R$ with $0 < a_1 < a_2 < \dots < a_n$,then what does $P(x)$ have?

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