Answer to Question #198329 in Calculus for Sanjeshni

Question

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.

Answer 1

a) Since f(x) = x⁴-2x³+2x-1, we have to derive it and equal that to zero to find the critical points:

$f(x)=x^4-2x^3+2x-1; f'(x)=4x^3-6x^2+2$

$f'(x)=4x^3-6x^2+2 =2(2x^3-3x^2+1) = 0$

To find the possible roots we try with the set given by $\frac{p}{q} = ±\{\frac{1}{1},\frac{1}{2}\}$, with this we find that f'(x) can be expressed as:

$f'(x)=2(2x^3-3x^2+1)=2(2x+1)(x-1)^2=0$

$x_1=+1; f(1)=0;x_2=-\frac{1}{2}; f(-\frac{1}{2})=-\frac{3}{16}$

(a) This means that the critical points are [ -1/2 , -3/16 ] and [ 1, 0 ].

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(b) To find the intervals where we'll check the following using f'(x) to know whether if increases or decreases:

$2(2x+1)(x-1)^2<0$

$2(2x+1)(x-1)^2>0$

With that information, we find how the slope of the tangent curve to f(x) is changing, after testing:

f'(x)<0 when (∞, -1/2)

f'(x)>0 when (-1/2, 1)

f'(x)>0 when (1, ∞)

(b) In conclusion, f(x) decreases on $x \in(\infin, - \frac{1}{2})$ and increases on $x \in(- \frac{1}{2},1)\, \land \,(1, \infin)$

Reference:

• Varberg, D. E., Purcell, E. J., & Rigdon, S. E. (2007). Calculus with differential equations. Pearson/Prentice Hall.