convex Explanation: To find that out, we need to get the second derivative first. Getting the first derivative. \displaystyle{\left[{1}\right]}\ \text{ }\ {f}'{\left({x}\right)}=\frac{{d}}{{\left.{d}{x}\right.}}{\left({x}^{{4}}-{4}{x}^{{3}}+{x}-{4}\right)} ...

As @André Nicolas suggests, the best approach is to factor by grouping. Factoring out x^3 from the terms in expression x^4-x^3 makes the left hand side of your equation become: x^3(x-1)+(x-1) ...

x4-4x2+x+2=0 Four solutions were found :  x = 1  x = -2  x =(1-√5)/2=-0.618  x =(1+√5)/2= 1.618 Step by step solution : Step  1  :Equation at the end of step  1  : (((x4) - 22x2) + x) + 2 = 0 ...

x4-4x3+2x-40=0 One solution was found :                         x ≓ -1.947012961 Step by step solution : Step  1  :Equation at the end of step  1  : (((x4) - 22x3) + 2x) - 40 = 0 Step  2  :Checking ...

It's easier to prove that \alpha\notin\mathbb{Q}(\beta), because this field is a subset of \mathbb{R}, but \alpha\notin\mathbb{R}. The factorization you found over \mathbb{R} shows the ...

Nghi N. · Kevin B. Apr 16, 2015 \displaystyle{f{{\left({x}\right)}}}={3}{x}^{{4}}-{4}{x}^{{3}}+{6} Derivative: \displaystyle{f}'{\left({x}\right)}={12}{x}^{{3}}-{12}{x}^{{2}}={12}{x}^{{2}}\cdot{\left({x}-{1}\right)} ...