Sketching a Graph Given Information about Its DerivativesDate: 07/30/2005 at 14:46:49 From: sahbina Subject: Graphing a function when not given f(x), but characteristics In class we are taking f'(x) and f''(x) and finding the intervals where f'(x) is increasing and decreasing and for f''(x) where the function is concave up and concave down, also any asymptotes. But for this question we are given some info and we have to graph the function. It's like going in reverse. For example, f(0)=4 and f(6)=0 f'(x)<0 if x<2 and x>4 f'(2) does not exist f'(4)=0 f'(x)>0 if 2<x<4 f''(x)<0, x cannot=2 From the information I can see that {0,6) are the boundaries. Also x=2,4 are points on the graph of f(x) because they make f'(x) and f''(x) equal 0. Altogether f(0), f(2), f(4), and f(6) are all points of the graph of f(x). My teacher told us f'(4)=0, but that is evident because 4 is an x value for f'(x) which makes it equal to 0. I attempted to graph the function, but I feel like there is a point missing on the graph because f(4) does not show up on the graph, and it is indeed a point. Also f(x) is decreasing at (-infinity, 2) and ( 4, +infinity), and f(x) is increasing at (2,4). Also f''(x) is concave down at (-infinity,2) and (2, +inifinity). Also there is a vertical asymptote at x=2. There has to be a point at f(4), but he didn't give it to us. Date: 07/30/2005 at 23:05:27 From: Doctor Peterson Subject: Re: Graphing a function when not given f(x), but characteristics Hi, Sahbina. I'm not sure what you mean by a few of your statements; your terminology is unclear sometimes. Be careful: x=2 is not a point, it's just a value of x! And f(0) is not a point, but a value of y. Let's look at the list of facts, and express them in English: >f(0)=4 and f(6)=0 The points (0,4) and (6,0) are on the graph >f'(x)<0 if x<2 and x>4 The graph is decreasing in the intervals (-oo,2) and (4,oo) >f'(2) does not exist The graph has no tangent, or a vertical tangent, at x=2 >f'(4)=0 The graph is horizontal at x=4 >f'(x)>0 if 2<x<4 The graph is increasing in (2,4) >f''(x)<0, x cannot=2 The graph is concave down everywhere except where its derivative does not exist There is no mention of boundaries; x can be anything. Rather, 0 and 6 are special values of x for which y is specified. Your comments on increasing, decreasing, and concave down intervals are correct, but you can't say there is a vertical asymptote at x=2, since then f(2) would not exist. (We would like to know f(2) and f(4), but since we aren't told those, we will just have to guess. It doesn't mean the points don't exist!) To sketch the graph, I would start by plotting the two specifically known points, (0,4) and (6,0). Then I would mark where the graph is increasing and decreasing, perhaps by making a dotted vertical line at x=2 and x=4 to mark the regions, and then drawing lightly a line slanting downward in the decreasing intervals: \| : : o : : |\ : /:\ + \ : / : \ | \ : / : \ + \ : / : \ | \ : / : \ + \ : / : \ | \:/ : \ --+---+---+---+---+---+---+---o---+---+--- | \ That zigzag doesn't represent the actual graph, just one fact about it, namely the direction of the slope. (I happen to have made it pass through the two given points, as well, which makes it fairly close to a possible graph of f.) Next, we can look at the details about the first derivative, namely that it is zero at x=4 and doesn't exist at x=2. We can draw a horizontal tangent at x=4, and (since we aren't told that f(4) doesn't exist) a vertical slope at x=2: \| : : o : : |\ : /:\ + \ : / : \ | \ : / : \ + \ : / : \ | \ : / : \ + \ : / : \ | \:/ : \ --+---+---+---|---+---+---+---o---+---+--- | \ Now we can check what is said about the second derivative: it's negative everywhere it exists. That means the graph is concave down. We have to "bend" the lines so they fit that description: * | : *** o : * : * | * : : * + * : * : * | * : : + *:* : * | : : + * : * | : : --+---+---+---+---+---+---+---o---+---+--- | I made sure it was always concave down, and (though you can't see it!) made it tangent to x=2 where it reaches it. That is called a cusp. None of the actual points except (0,4) and (6,0) are necessarily correct, and the local minimum and maximum may be far from where they really belong; but we've sketched a graph that fits all the information we were given, which is the goal. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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