The function f(x) = 4x + 1 is a line, therefore it's continuous everywhere. In particular, it's continuous at x = 5 with f(5) = 21. How must the x-values be restricted if we want to have 20.5< f(x) < 21.5?
Since we're looking at continuity near x = 5, 5 is our value of c and f(c) = f(5) = 21. Graph the function and set the window so that
0 ≤ x ≤ 10 and 0 ≤ y ≤ 42.
We do this so that (5, 21) is in the center of the window. Now start narrowing the values of x, keeping 5 in the center, until the function goes out the sides of the graph instead of the top and bottom. Here's one possible progression, where we first bring x within 1 step of 5, then within 0.5 of 5, then within 0.25 of 5, and finally within 0.125 of 5:
If we restrict x so that 4.875 ≤ x ≤ 5.125,we find what we want: a picture where (5, 21) is in the middle, and all the values of f(x) are between 20.5 and 21.5.
is discontinuous at x = 1. How does this function fail the continuous function guarantee?
If we try to get f(x) within 0.5 of f(1) = 2 (that is, 1.5 < f(x) < 2.5), we run into trouble. No matter how much we restrict x, since we aren't allowed to have 1 ≤ x ≤ 1, the picture will always have some values of x that are less than 1. Since f(x) will be less than 1 for these values, f(x) will not be between 1.5 and 2.5. We just can't get the picture we want. Sniff.
For the given function, value of c, and specified range of f(x), find an appropriate restriction of x that produces the kind of picture we want.
Remember that there are multiple right answersfor each of these. As long as our answer produces the correct sort of graph, it's fine.
2.8 ≤ x ≤ 3.2 will work, as will anything more restricted. (We can require x be 0.2 away from 3, or closer, but farther away won't work.)
The largest range of values that will work here is ln(0.9) ≤ x < ≤ ln(1.1). We know this because eln(0.9) = 0.9 and eln(1.1) = 1.1.
Trial and error is best here. Tinker with some x-values until you find a range that works. One possible range of values is -2.1 ≤ x ≤ -1.9.