Question

what best approximates the intervals on which the graph of the function f(x)=x^4+2x^3-5x^2-6x is decreasing?

Answer 1

Solution: The function
`f(x)=x^4+2x^3-5x^2-6x` is decreasing on
`(-\infty ,-2.303]\cup [-0.5,1.303]`.

Step-by-step explanation:

The given function is
`f(x)=x^4+2x^3-5x^2-6x`.

differentiate with respect to x.

`f'(x)=4x^3+6x^2-10x-6`

To find the extreme points put
`f'(x)=0` and find the value of x.

`f'(x)=4x^3+6x^2-10x-6=0`

`2(2x^3+3x^2-5x-3)=0`

By hidden trial
`(-1)/(2)` is the root of the equation and
`(x+(1)/(2) )` is factor of the equation.

`2(x+(1)/(2) )(2x^2+2x-6)=0\\4(x+(1)/(2) )(x^2+x-3)=0\\4(x+(1)/(2) )(x+2.303)(x-1.303)=0`

The extreme points of graph
`x=-0.5, -2.303, 1.303`

The intervals are
`(-\infty ,-2.303]\cup[-2.303,-0.5]\cup [-0.5,1.303]\cup [1.303,\infty )`.

Put any value from each interval in
`f'(x)=4x^3+6x^2-10x-6`.

If the value of
`f'(x)>0`, then the function increase on that interval.

If the value of
`f'(x)<0`, then the function decrease on that interval.

Since the value
`f'(x)<0` on
`(-\infty ,-2.303]`. and
`[-0.5,1.303]`, therefore function
`f(x)=x^4+2x^3-5x^2-6x` is decreasing on
`(-\infty ,-2.303]\cup [-0.5,1.303]`.