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It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if
</span><span>
</span><span>In notation we write respectively
</span>
Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²<span>
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c</span>²
That is to say, if c = -2, f(x) is continuous at x = 4.
Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers
The translation of the given figure will have vertices, <u>A'(1,2),</u> <u>B'(3,2),</u> <u>C'(3,-5),</u> and <u>D(-1,-3)</u> from the initial vertices A(3,9), B(5,9), C(5,2), and D(1,4). The figure is attached.
In the question, we are asked to translate the given figure 2 units left and 7 units down.
In the translation, the left-right movement implies the change in the x-coordinate, whereas the up-down movement implies the change in the y-coordinate.
For any point P(x, y) the given rule says that after translation,
P(x, y) becomes P'(x - 2, y - 7).
The coordinates of the given figure are:
A(3,9), B(5,9), C(5,2), and D(1,4).
After translation, these points can be shown as:
A(3,9) becomes A'(1,2).
B(5,9) becomes B'(3,2).
C(5,2) becomes C'(3,-5).
D(1,4) becomes D'(-1,-3).
Thus, the translation of the given figure will have vertices, <u>A'(1,2),</u> <u>B'(3,2),</u> <u>C'(3,-5),</u> and <u>D(-1,-3)</u> from the initial vertices A(3,9), B(5,9), C(5,2), and D(1,4). The figure is attached.
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