Determine the monotonic intervals of the functions

calculusfunctions

i need to determine the monotonic intervals of this function $y=2x^3-6x^2-18x-7$. I tried the below but i am not sure if i am doing it right.

My work: \begin{align} y=2x^3-6x^2-18x-7
&\Longleftrightarrow 6x^2-12x-18=0\\
&\Longleftrightarrow6(x^2-2x-3)=0\\
&\Longleftrightarrow(x-3) (x+1)\\
&\Longleftrightarrow x-3=0 x+1=0\\
&\Longleftrightarrow x=3, x=-1\\
\end{align}

so my function increases when
$x\in[3, +\infty[$ and decreases when $x\in[-1, 3]\cup ]-\infty, -1]$

Please i want to know how to solve this problem any help with explanation will be appreciated. thanks in advanced

Best Answer

You have correctly found the derivative

$$\frac{dy}{dx}=6x^2-12x-18=6(x^2-2x-3)=6(x-3)(x+1)$$

and where it is zero, but you have not quite got the intervals correct.

The derivatives is positive if and only if

$$(x-3)(x+1)>0$$

which is positive if and only if

$$(x>3 \text{ and } x>-1)\quad \text{ or }\quad (x<3 \text{ and } x<-1)$$

i.e. if and only if

$$(x>3)\quad \text{ or }\quad (x<-1)$$

So the function is (strictly) increasing on the interval $(-\infty, -1]$ and on the interval $[3,\infty)$.

The function is (strictly) decreasing on the interval $[-1,3]$.

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Finding first and second derivatives of $y=(x-1) (x+1)^3$ and $y=\frac{10}{4x^3-9x^2+6x}$ in order to determine monotonic intervals

I'm having a problem finding the first and second derivatives of the following...

Hint: I will show the first derivative of each function below, hope this helps:

for $y=(x-1) (x+1)^3$

you could use the chain rule: $\frac{d}{dx}(u.v)=v.\frac{d}{dx}(u)+u.\frac{d}{dx} (v)$

Choose $$u=x-1, v=(x+1)^3$$

$$\frac{d}{dx}(u)=\frac{d}{dx}(x-1)=1$$ $$\frac{d}{dx}(v)=\frac{d}{dx}(x+1)^3=3(x+1)^{3-1}(x-1)=3(x+1)^2(x-1)$$ $$\frac{d}{dx} (x-1) (x+1)^3 =(x+1)^3+(x-1)(3(x+1)^2)$$

Now for the function: $y=\dfrac{10}{4x^3-9x^2+6x}$ This function is of the form $g=\frac{u}{v}$ and the derivative is:

$$\frac{\frac{d}{dx}(u).v -u\frac{d}{dx}(v)}{v^2}$$ choose $$u=10, v= 4x^3-9x^2+6x$$ $$\frac{d}{dx}(u)=\frac{d}{dx}(10)=0$$ $$\frac{d}{dx}(v)=\frac{d}{dx} (4x^3-9x^2+6x)=12x^2-18x+6$$

$$\frac{d}{dx}(\frac{10}{4x^3-9x^2+6x})=\frac{-10(12x^2-18x+6)}{(4x^3-9x^2+6x)^{2}}$$

The above derivative is not defined where v=0, that is at values:

$$x=0 , x=\frac{9}{8}+i\frac{\sqrt {15}}{8}, \frac{9}{8}-i\frac{\sqrt {15}}{8}$$

Best Answer

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