# Increasing or Decreasing?

Before we continue with applications of derivatives, including optimization and curve sketching, we will want to be able to determine when a function is increasing/decreasing and the intervals of concavity. As we work with functions, knowing the shape of the graph will help us determine what points of interest there are in different situations. We begin here by determining the intervals increasing and decreasing and local extrema. In our next post, we will look at concavity.

## Determine the shape of a function

Here we will be working with the function
\begin{align*}
f(x)=x(x-2)^{3}.
\end{align*}
With this function we will want to find

• the critical points,
• intervals of increasing and decreasing,
• local extrema.

As we work through this problem, we will work one step at a time.

### First Derivative

In order to determine the intervals of increasing and decreasing as well as critical points and extrema, we will be working with the first derivative of the function. We will, therefore, need to find this. As we look at
\begin{align*}
\frac{d}{dx}x(x-2)^{3}
\end{align*}
we note that we first have a product rule. Here the first function is $$F=x$$ and the second is $$S=(x-2)^{3}$$. The derivative of the first is $$F’=1$$, while the derivative of the second requires a chain rule. Here the inside function is $$i=(x-2)$$ and the outside function is $$o=i^{3}$$. Combining these together, we get
\begin{align*}
\frac{d}{dx}x(x-2)^{3}&=SF’+FS’ \\
&=(x-2)^{3}(1)+x(3(x-2)^{2}) \\
&=(x-2)^{2}((x-2)+3x) \\
&=(x-2)^{2}(4x-2).
\end{align*}

While I could have gone through this in more detail, I will refer you to the post Derivative Examples if you would like more practice with taking derivatives. Further note, that we have taken the time to simplify the derivative because we will be using it.

#### Finding zeroes

Now that we have the derivative of $$f(x)$$, we will need to determine the points where the derivative is undefined or $$0$$. These points are important because they are the potential critical points of the function. Furthermore, even if they are not critical points, these points are still important because they allow us to partition the number line into separate intervals over which the function is either increasing or decreasing, but not both. We then find that
\begin{align*}
f'(x)&=0 \\
(x-2)^{2}(4x-2)&=0 \\
(x-2)^{2}&=0 \text{ or } 4x-2=0 \\
x&=2 \text{ or } x=\frac{1}{2}.
\end{align*}

#### Finding Positive and Negative

We can now focus on numbers in the intervals $$(-\infty,\frac{1}{2})$$, $$(\frac{1}{2},2)$$ and $$(2,\infty)$$. Since the derivative is not $$0$$ or undefined on these intervals, it cannot change from positive to negative. Therefore, if we know what happens at one of the points, we know what happens at all of them. We now find

• Choose $$x< \frac{1}{2}$$, say $$x=0$$.
• We want to know if the derivative is positive or negative, so we find $$f'(0)=-8$$.
• Since $$f'(0)=-8 <0$$, we know $$f'(x) < 0$$ for all $$x$$ in $$(-\infty, \frac{1}{2})$$.
• Choose $$x$$ with $$\frac{1}{2} < x< 2$$, say $$x=1$$.
• We want to know if the derivative is positive or negative, so we find $$f'(1)=2$$.
• Since $$f'(1)=2 >0$$, we know $$f'(x) > 0$$ for all $$x$$ in $$(\frac{1}{2}, 2)$$.
• Choose $$x> 2$$, say $$x=3$$.
• We want to know if the derivative is positive or negative, so we find $$f'(3)=10$$.
• Since $$f(0)=10 >0$$, we know $$f'(x) > 0$$ for all $$x$$ in $$(2, \infty)$$.

We can summarize this with the number line below. #### Summarizing

Now that we’ve determined where the first derivative is $$0$$, undefined, positive and negative, we can now answer the questions about increasing, decreasing, critical points and extrema.

• Since the derivative is $$0$$ at $$x=\frac{1}{2}$$ and $$x=2$$, we check these to see if they are in the domain.
• Both of these are in the domain, so we have critical values at $$x=\frac{1}{2}$$ and $$x=2$$.
• In order to find our critical points, we also need to the $$y$$-values for the critical values.
• We get that $$f(\frac{1}{2})=\frac{1}{2}(\frac{1}{2}-2)^{3}=\frac{-27}{16}$$.
• Furthermore, $$f(2)=2(2-2)^{3}=0$$.
• Therefore, we have critical points at $$(\frac{1}{2},\frac{-27}{16})$$ and $$(2,0)$$.
• Note the derivative is negative on the interval $$(-\infty,\frac{1}{2})$$, and positive on $$(\frac{1}{2},2) \cup (2, \infty)$$.
• This gives us that he interval of increasing is $$(\frac{1}{2},\infty)$$ and the interval of decreasing is $$(-\infty,\frac{1}{2})$$.
• Because the function changes from decreasing to increasing at $$x=\frac{1}{2}$$, we will have a local minimum at $$x=\frac{1}{2}$$.
• The local minimum will be $$f(\frac{1}{2})=\frac{-27}{16}$$.

As a note, we would say the function is increasing over the entire interval $$(\frac{1}{2},\infty)$$ instead of $$(\frac{1}{2},2) \cup (2,\infty)$$ because the derivative is positive over this interval except at the single point at $$x=2$$ where the function is still defined. If we had that the function was not defined at $$2$$, we would have broken this interval up. Furthermore, if the derivative had been $$0$$ for an interval larger than a single point, we would have broken off this interval and said the function was constant for this interval. However, depending on how your instructor presents this, they may ask for different things at these points. Therefore, I would suggest following whatever your instructor asks of you at these points.

## Conclusion

Here we have started to find the shape of the graph by determining where the function is increasing, decreasing or has extrema. In our next post, we will look at the concavity of the graph so that we can get a more complete picture of what is happening with the graph.

As always, I hope this helped with your studies and that you enjoyed the process. Make sure to like this post or share on Social Media if you found it helpful. Also, check back to make sure you can see the analysis of the concavity of the function.

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