Calculus: Early Transcendentals 8th Edition

(a) $f$ is increasing on 2 open intervals: $(1, 3)$ and $(4,6)$. (b) $f$ is decreasing on 2 open intervals: $(0, 1)$ and $(3,4)$. (c) On interval $(0,2)$, $f$ is concave upward. (d) On intervals $(2,4)$ and $(4,6)$, $f$ is concave downward. (e) Point $(2,3)$ is the only point of inflection..
(a) To find the open intervals on which $f$ is increasing by looking at the graph, we look for all the places on the graph where the line has an UPWARD trend, then we look down on the $Ox$ line to find out which interval it is. Here, we find that from $x=1$ to $x=3$, the graph goes up from $y=2$ to $y=4$. Also, from $x=4$ to $x=6$, the graph goes up from $y=1$ to $y=6$. Therefore, $f$ is increasing on 2 open intervals: $(1, 3)$ and $(4,6)$. (b) To find the open intervals on which $f$ is decreasing by looking at the graph, we look for all the places in the graph where the line has a DOWNWARD trend, then we look down on the $Ox$ line to find out which interval it is. Here, we find that from $x=0$ to $x=1$, the graph goes down from $y=5$ to $y=2$. Also, from $x=3$ to $x=4$, the graph goes down from $y=4$ to $y=1$. Therefore, $f$ is decreasing on 2 open intervals: $(0, 1)$ and $(3,4)$. (c) 'Concave upward' is where the graph SAGS DOWNWARD. So it would create the sight of a bag, whose depth depends on the extremity of the sagging. Also, for more detail, by definition, 'concave upward' is where the graph $f$ lies ABOVE all of its tangents on an interval I. Taking all those facts, we find that in this graph, on the interval $(0,2)$, $f$ is concave upward. (The interval $(2,3)$ is in fact a part where the graph is concave downward) (Also, it might be tempting to state on interval $(3,6)$, $f$ is also concave upward. This is unfortunately not true, since all the tangents here lie above graph $f$) (d) 'Concave downward' is where the graph "SAGS" UPWARD. So it would create the sight of an upside down bag, whose depth depends on the extremity of the "sagging". Also, for more detail, by definition, 'concave upward' is where the graph $f$ lies BELOW all of its tangents on an interval I. Taking all those facts, we find that in this graph, on the intervals $(2,4)$ and $(4,6)$, $f$ is concave downward. (We cannot combine the 2 intervals together since there is no tangent line at $x=4$. In other words, $f$ is not differentiable at $x=4$.) (e) According to the definition, a point P on a curve $y=f(x)$ is called an inflection point if: - $f$ is continuous at the point $P$. - The curve changes from concave upward to downward or vice versa at $P$. Therefore, we need to look back again at parts c) and d) to find the place where the curve changes its concavity. On $(0,2)$, $f$ is concave upward. On $(2,4)$, $f$ is concave downward. So at point where $x=2$, it changes concave direction. That means point $(2,3)$ is the only inflection point here.