The range of the polynomial P(x) = 4x^3 - 3x | vedclass

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Question

(C) Given the polynomial $P(x) = 4x^3 - 3x$ for $x \in \left(-\frac{1}{2}, \frac{1}{2}\right)$.To find the range,we analyze the derivative $P'(x) = 12

Vedclass · Accepted Answer

$(-1, 1)$

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(C) Given the polynomial $P(x) = 4x^3 - 3x$ for $x \in \left(-\frac{1}{2}, \frac{1}{2}\right)$.To find the range,we analyze the derivative $P'(x) = 12x^2 - 3 = 3(4x^2 - 1) = 3(2x - 1)(2x + 1)$.For $x \in \left(-\frac{1}{2}, \frac{1}{2}\right)$,we have $4x^2 Thus,$P'(x) Since $P(x)$ is continuous and strictly decreasing,the range is $(P(1/2), P(-1/2))$.Calculating the values at the boundaries:$P(1/2) = 4(1/8) - 3(1/2) = 1/2 - 3/2 = -1$.$P(-1/2) = 4(-1/8) - 3(-1/2) = -1/2 + 3/2 = 1$.Since the interval is open,the range is $(-1, 1)$.

The range of the polynomial $P(x) = 4x^3 - 3x$ as $x$ varies over the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is

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