The number of distinct real roots of the equa | vedclass

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(C) Let $f(x) = 3x^{4} + 4x^{3} - 12x^{2} + 4$.To find the critical points,we find the derivative:$f'(x) = 12x^{3} + 12x^{2} - 24x = 12x(x^{2} + x - 2

Vedclass · Accepted Answer

$4$

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(C) Let $f(x) = 3x^{4} + 4x^{3} - 12x^{2} + 4$.To find the critical points,we find the derivative:$f'(x) = 12x^{3} + 12x^{2} - 24x = 12x(x^{2} + x - 2) = 12x(x + 2)(x - 1)$.The critical points are $x = -2, 0, 1$.Now,evaluate $f(x)$ at these critical points:$f(-2) = 3(-2)^{4} + 4(-2)^{3} - 12(-2)^{2} + 4 = 3(16) + 4(-8) - 12(4) + 4 = 48 - 32 - 48 + 4 = -28$.$f(0) = 3(0)^{4} + 4(0)^{3} - 12(0)^{2} + 4 = 4$.$f(1) = 3(1)^{4} + 4(1)^{3} - 12(1)^{2} + 4 = 3 + 4 - 12 + 4 = -1$.As $x 	o \infty$,$f(x) 	o \infty$ and as $x 	o -\infty$,$f(x) 	o \infty$.Analyzing the sign changes:$1$. In $(-\infty, -2)$,$f(x)$ decreases from $\infty$ to $-28$. Since $f(-2) = -28 $2$. In $(-2, 0)$,$f(x)$ increases from $-28$ to $4$. Since $f(-2)  0$,there is one root in $(-2, 0)$.$3$. In $(0, 1)$,$f(x)$ decreases from $4$ to $-1$. Since $f(0) > 0$ and $f(1) $4$. In $(1, \infty)$,$f(x)$ increases from $-1$ to $\infty$. Since $f(1) Thus,there are $4$ distinct real roots.

The number of distinct real roots of the equation $3x^{4} + 4x^{3} - 12x^{2} + 4 = 0$ is ..... .

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