3.3 Rates of change and behavior of graphs  (Page 4/15)

We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from $t=1$ to $t=3$ and from $t=4$ on.

In interval notation    , we would say the function appears to be increasing on the interval (1,3) and the interval $(4,\infty ).$

Using technology, we find that the graph of the function looks like that in [link] . It appears there is a low point, or local minimum, between $x=2$ and $x=3,$ and a mirror-image high point, or local maximum, somewhere between $x=\mathrm{-3}$ and $x=\mathrm{-2.}$

Observe the graph of $f.$ The graph attains a local maximum at $x=1$ because it is the highest point in an open interval around $x=1.$ The local maximum is the $y$ -coordinate at $x=1,$ which is $2.$

The graph attains a local minimum at $x=\mathrm{-1}$ because it is the lowest point in an open interval around $x=\mathrm{-1.}$ The local minimum is the y -coordinate at $x=\mathrm{-1},$ which is $\mathrm{-2.}$

Observe the graph of $f.$ The graph attains an absolute maximum in two locations, $x=\mathrm{-2}$ and $x=2,$ because at these locations, the graph attains its highest point on the domain of the function. The absolute maximum is the y -coordinate at $x=\mathrm{-2}$ and $x=2,$ which is $16.$

The graph attains an absolute minimum at $x=3,$ because it is the lowest point on the domain of the function’s graph. The absolute minimum is the y -coordinate at $x=3,$ which is $\mathrm{-10.}$