Answer
(a) $f$ is increasing on 3 open intervals: $(0, 1)$, $(3,5)$ and $(5,7)$.
(b) $f$ is decreasing on 1 open interval: $(1,3)$.
(c) On 2 intervals $(2,4)$ and $(5,7)$, $f$ is concave upward.
(d) On 2 intervals $(0,2)$ and $(4,5)$, $f$ is concave downward.
(e) 3 points $(2,2)$, $(4,3)$ and $(5,4)$ are points of inflection.
Work Step by Step
(a)
To find the open intervals on which $f$ is increasing by looking at the graph, we look for all the places in 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 $x=0$ to $x=1$, the graph goes up from $y=0$ to $y=3$. Also, from $x=3$ to $x=5$, the graph goes up from $y=1$ to $y=4$. Finally, from $x=5$ to $x=7$, the graph goes up from $y=4$ to $y=6$. (At $x=5$, the graph does not increase or decrease.)
Therefore, $f$ is increasing on 3 open intervals: $(0, 1)$, $(3,5)$ and $(5,7)$.
(b)
To find the open intervals on which $f$ is decreasing again 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 only from $x=1$ to $x=3$, the graph goes down from $y=3$ to $y=1$.
Therefore, $f$ is decreasing on 1 open interval: $(1,3)$.
(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 interval $(2,4)$, $f$ is concave upward. Another interval not so obvious where $f$ is also concave upward is $(5,7)$ (it has only the sight of half a bag, but still all the tangents there lie below $f$).
(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 downward' 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 intervals $(0,2)$ and $(4,5)$, $f$ is concave downward. (Again, the interval $(4,5)$ is not so obvious but still all the tangents there lie above $f$.)
(e)
According to definition, a point P on a curve $y=f(x)$ is called an inflection point if:
- $f$ is continuous at 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 downward. Then on $(2,4)$, $f$ is concave upward. Next, on $(4,5)$, $f$ is concave downward. Finally, on $(5,7)$, $f$ is concave upward.
So the points where $x=2$, $x=4$ and $x=5$ are those we are looking for. That means points $(2,2)$, $(4,3)$ and $(5,4)$ are points of inflection.
