The number of distinct real roots of x^{4}-4x | vedclass

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(D) Let $f(x) = x^{4}-4x+1$.Find the derivative: $f'(x) = 4x^{3}-4$.Set $f'(x) = 0$ to find critical points: $4x^{3}-4 = 0 \Rightarrow x^{3} = 1 \Righ

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$2$

Vedclass · Answer

(D) Let $f(x) = x^{4}-4x+1$.Find the derivative: $f'(x) = 4x^{3}-4$.Set $f'(x) = 0$ to find critical points: $4x^{3}-4 = 0 \Rightarrow x^{3} = 1 \Rightarrow x = 1$.Since $f''(x) = 12x^{2}$,$f''(1) = 12 > 0$,so $x = 1$ is a point of local minima.The minimum value is $f(1) = 1^{4}-4(1)+1 = 1-4+1 = -2$.As $x 	o \infty$,$f(x) 	o \infty$ and as $x 	o -\infty$,$f(x) 	o \infty$.Since the minimum value is $-2$ (which is less than $0$) and the function is continuous,the graph crosses the $x$-axis at two distinct points: one in the interval $(-\infty, 1)$ and one in the interval $(1, \infty)$.Therefore,there are $2$ distinct real roots.

The number of distinct real roots of $x^{4}-4x+1=0$ is

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