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 $(-\infty ,-2.303]\cup [-0.5,1.303]$.

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 $\frac{-1}{2}$ is the root of the equation and $(x+\frac{1}{2} )$ is factor of the equation.

$2(x+\frac{1}{2} )(2x^2+2x-6)=0\\4(x+\frac{1}{2} )(x^2+x-3)=0\\4(x+\frac{1}{2} )(x+2.303)(x-1.303)=0$

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 $f'(x)$, then the function decrease on that interval.

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