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While analyzing a given function, there are features we may need to define. Whether the function is increasing or decreasing and its highest and lowest points are just a few of the items we can find using the power of derivatives.

Let’s start off by considering the following equation over an interval [4,4]:

$f(x)=3x^4-4x^3-12x^2+3$

The first thing we realize when looking at this equation is that it would be brutal to graph by hand. However, using the rules of the derivative, we can determine all that we need to know without even seeing it graphically.

To determine if a function is increasing or decreasing over a given interval (I), consider the following:

If $f’(x)>0$ for all ${x\in I}$, then the function is increasing over the interval

If $f’(x)<0$ for all ${x\in I}$, then the function is decreasing over the interval

#### Example

Determine where the function is increasing and decreasing:

$f(x)=3x^4-4x^3-12x^2+3$

First determine $f’(x)$:

$f(x)=12x^3-12x^2-24x$

$\begin{tabular}{|l|l|} \hline x&f(x) \\ \hline -3.5 & -577 \\ \hline \hline -3 & -360 \\ \hline -2.5& -202 \\ \hline \hline -2 & -96 \\ \hline -1.5 & -31 \\ \hline \hline -1& 0 \\ \hline -.5& 7 \\ \hline \hline 0 &0 \\ \hline .5 & -13\\ \hline \hline 1 & -24 \\ \hline 1.5 &-22 \\ \hline \hline 2 & 0 \\ \hline 2.5 & 52 \\ \hline \hline 3 &144 \\ \hline 3.5 & 283 \\ \hline \end{tabular}$

With these values in the table, we can confirm that the function is decreasing on the interval $(-4,-1) \in(0,2)$. The function is increasing on the interval $(-1,0) \in (2,4)$.

Using the derivative, we can also determine where the function reaches critical points ${(x_{{0}})}$, in which a maxima or a minima is present. These critical points occur ${{f}^{‘}(x)=0}$. To determine if the critical point is a maxima or a minima, the following rules can be applied:

If ${{f}^{‘}(x)}>0$ on an open interval to the left of the critical point, ${x_{{0}}}$, and ${{f}^{‘}(x)}<0$ on an open interval to the right of ${x_{{0}}}$, then the function is at a relative maxima at ${x_{{0}}}$.

If ${{f}^{‘}(x)}<0$ on an open interval to the left of the critical point, ${x_{{0}}}$, and ${{f}^{‘}(x)}>0$ on an open interval to the right of ${x_{{0}}}$, then the function is at a relative minima at ${x_{{0}}}$.

If ${{f}^{‘}(x)}$ has the same sign on either side of ${x_{{0}}}$, then the function is not at a relative maxima or minima at ${x_{{0}}}$.

Concavity of a function is another characteristic that can be determined using the derivative. Concavity tells us whether the function is open upward or downward over a given interval and is determined using the second derivative of the given function:

If ${{f}^{”}(x)>0}$ for all ${x\in I}$, then the function is concave upward over the interval

If ${{f}^{”}(x)<0}$ for all ${x\in I}$, then the function is concave downward over the interval

#### Example

To illustrate this, the second derivative of the function must first be taken. Continuing from above, we find

${{f}^{”}(x):}$

\begin{equation*}

{{f}^{”}(x)=\text{36}x^{{2}}-\text{24}x-\text{24}}

\end{equation*}

$\begin{tabular}{|l|l|} \hline x&f(x) \\ \hline -3.5 & 501 \\ \hline \hline -3 & 372 \\ \hline -2.5& 261 \\ \hline \hline -2 & 168 \\ \hline -1.5 & 93 \\ \hline \hline -1& 36 \\ \hline -.5& -3 \\ \hline \hline 0 &-24 \\ \hline .5 & -27\\ \hline \hline 1 & -12 \\ \hline 1.5 &21 \\ \hline \hline 2 & 72 \\ \hline 2.5 &141 \\ \hline \hline 3 &228 \\ \hline 3.5 & 333 \\ \hline \end{tabular}$

With these values in the table, we can confirm that the function is concave upward on the interval $(-4,-1) \in (1.5,4)$. The function is concave downward on the interval (-.5,1).

Where ${{f}^{”}(x)=0}$ there exists an inflection point, where the concavity of the curve changes from either upward to downward or downward to upward.