Question #201011

Conside the function f(x)=x to the power 4-2xcube+2x-1

A. Find the critical points of f(x)

B. Determine the interval over which f(x) is increasing and the interval on which it is decreasing.


Expert's answer

Let us consider the function f(x)=x4−2x3+2x−1f(x)=x^4-2x^3+2x-1.


A. Let us find the critical points of f(x)f(x). The funtion is differentiable in all points of the real line. Since f′(x)=4x3−6x2+2f'(x)=4x^3-6x^2+2, we conclude that 4x3−6x2+2=04x^3-6x^2+2=0 implies 2(x−1)(2x2−x−1)=02(x-1)(2x^2-x-1)=0, and hence 2(x−1)2(2x+1)=02(x-1)^2(2x+1)=0. Therefore, x1=1x_1=1 and x2=−0.5x_2=-0.5 are critical points of the function f(x).f(x).


B. Let us determine the interval over which f(x)f(x) is increasing and the interval on which it is decreasing. Taking into account that (x−1)2≥0(x-1)^2\ge0, we conclude that for x>−0.5x>-0.5 we have that f′(x)=2(x−1)2(2x+1)≥0f'(x)=2(x-1)^2(2x+1)\ge 0, and hence the function f(x)f(x) is increasing on the interval (−0.5,+∞)(-0.5, +\infty). For x<−0.5x<-0.5 we have that f′(x)=2(x−1)2(2x+1)<0f'(x)=2(x-1)^2(2x+1)< 0, and hence the function f(x)f(x) is decreasing on the interval (−∞,−0.5).(-\infty, -0.5).



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