The function f(x) is defined on [-6, 6] and has the graph:
Difficulty Rating: 
(a) The blue graph shows that no negative output values are possible in y = | f(x) |
(b) The yellow graph shows that y = f( |x| ) makes positive and negative inputs the same, so an input of 3 must yield the same value as an input of -3. The graph automatically becomes y-symmetric, and the original domain values less than zero are replaced.
(c) The green graph shows -f(x), the graph of f(x) reflected across the x-axis.
(d) The orange graph shows f(x/2); note that the graph is effectively "stretched" horizontally. In fact, each input value occurs twice as far down the x-axis. This is the equivalent of a graph stretching twice as high by graphing 2f(x). However, when you are manipulating the input, the graph is affected horizontally rather than vertically, and the effect is the opposite of what you might think (1/2 actually stretched the graph twice as large rather than shrinking it half as small).
(e) The purple graph is the original function moved one unit to the right, as demanded by f(x-1). Note again that affecting the input moves the graph in a direction opposite to what you might reason: a "-1" moves the graph to the right, rather than one to the left.
At 8:00 am on Saturday, a man
begins running up the side of a mountain to his weekend campsite. On Sunday
morning at 8:00 am, he runs back down the mountain. It takes him 20 minutes
to run up, but only 10 minutes to run down. At some point on his way down,
he realizes that he passed the same place at exactly the same time on Saturday.
Prove that he is correct.
(Hint: Let u(t) and d(t) be the position functions
for the runs up and down, and apply the Intermediate Value Theorem to f(t)
= u(t) - d(t).
Difficulty Rating: 
Source: Calculus, 6ed., by Larson, Hostetler, and Edwards,
p. 78
Remember that u and d are position functions; they give us the runner's height relative to the ground. So, if the runner is 300 feet off the ground at t = 5 minutes on the way up, u(5)=300. We'll say that his campsite is some fixed height (which it must be) and let's just say for fun that the height is 2000 feet. (This number is not important; it just helps ye olde thinkinge processe).
We know that u(0)--the starting time--equals 0, as he has not yet left the ground, and u(20)=2000. We also know that d(0)=2000 and d(10)=0. Let's think about values of f.We can see that f(0)=u(0) - d(0)= 0 - 2000 = -2000. Also, f(20) = 2000 - 0 = 2000. If you think about it, d(c) = 0 if c is 10 or greater, as after 10 minutes have passed, our runner is already finished descending the mountain. By the very nature of physical laws, u must be a continuous function--he cannot suddenly jump from 2 feet to 25 feet off the ground without passing all the distances in the middle along the way. There are no time-space wormholes on mountainsides--just wormholes made by real life worms. By the same reasoning, d must also be a continuous function. Therefore, f must be a continuous function, as it is the result of subtracting two continuous functions. This is good. Really good. (And we're almost finished.) The fact that f is continuous guarantees that from f(0) to f(20), the function must "cover" all outputs from -2000 to 2000, as demonstrated in the below picture, and as promised by the Intermediate Value Theorem.
You may disagree with what the graph looks like, but one thing is for sure--it must cross the x-axis. There must be a time where f(t) = 0, which must mean that u(t) and d(t) were exactly the same! So, at some point t, the position going up was exactly the same as the position going down!
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