Solution: The function [tex]f(x)=x^4+2x^3-5x^2-6x[/tex] is decreasing on [tex](-\infty ,-2.303]\cup [-0.5,1.303][/tex].
Explanation:
The given function is [tex]f(x)=x^4+2x^3-5x^2-6x[/tex].
differentiate with respect to x.
[tex]f'(x)=4x^3+6x^2-10x-6[/tex]
To find the extreme points put [tex]f'(x)=0[/tex] and find the value of x.
[tex]f'(x)=4x^3+6x^2-10x-6=0[/tex]
[tex]2(2x^3+3x^2-5x-3)=0[/tex]
By hidden trial [tex]\frac{-1}{2}[/tex] is the root of the equation and [tex](x+\frac{1}{2} )[/tex] is factor of the equation.
[tex]2(x+\frac{1}{2} )(2x^2+2x-6)=0\\4(x+\frac{1}{2} )(x^2+x-3)=0\\4(x+\frac{1}{2} )(x+2.303)(x-1.303)=0[/tex]
The extreme points of graph [tex]x=-0.5, -2.303, 1.303[/tex]
The intervals are [tex](-\infty ,-2.303]\cup[-2.303,-0.5]\cup [-0.5,1.303]\cup [1.303,\infty )[/tex].
Put any value from each interval in [tex]f'(x)=4x^3+6x^2-10x-6[/tex].
If the value of [tex]f'(x)>0[/tex], then the function increase on that interval.
If the value of [tex]f'(x)<0[/tex], then the function decrease on that interval.
Since the value [tex]f'(x)<0[/tex] on [tex](-\infty ,-2.303][/tex]. and [tex][-0.5,1.303][/tex], therefore function [tex]f(x)=x^4+2x^3-5x^2-6x[/tex] is decreasing on [tex](-\infty ,-2.303]\cup [-0.5,1.303][/tex].
