Show that for any real numbers q, the polynomial P(x) = x^7 + x^3 + q , has exactly one real root.

Question

Sher Mohammad · Answer

P'(x)=7x^6+3x^2

P'(x)=0

x^2(7x^4+2)=0, imply x=0

since f'(x)=0 at only x=0 and for any other x f'(x)>0, only one point of inflection possible for p(x) imply exact one root.

Sher Mohammad

B.Tech, IIT Delhi

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Show that for any real numbers q, the polynomial P(x) = x^7 + x^3 + q , has exactly one real root.

Show that for any real numbers q, the polynomial P(x) = x^7 + x^3 + q , has exactly one real root.

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Show that for any real numbers q, the polynomial P(x) = x^7 + x^3 + q , has exactly one real root.

Show that for any real numbers q, the polynomial P(x) = x^7 + x^3 + q , has exactly one real root.

1 Answers

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Other Related Questions on Differential Calculus

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