Given that $y = x \sqrt{1 - 4x^2}$ , find the range of values of $x$ for $y$ to be v

Question

Given that $y = x \sqrt{1 - 4x^2}$ , find the range of values of $x$ for $y$ to be valid, and for $-1/2<x<1/2$, prove that $y \frac{\mathrm{d}y}{\mathrm{d}x} + x(8x^2 - 1) = 0$.

Accepted Answer

(i) $-\dfrac12\le x\le\dfrac12$ (ii) $y\dfrac{\mathrm{d}y}{\mathrm{d}x}+x(8x^2-1)=0$

Quadratic Functions, Equations and Inequalities - Quadratic Inequalities

Quadratic Functions, Equations and Inequalities - Quadratic Inequalities

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