AP Calculus
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Special Focus: The Fundamental<br />
Theorem of <strong>Calculus</strong><br />
(c) This question will be one of the most challenging for students because the<br />
composite function is created by the upper limit. The domain of f is 0 ≤ x ≤ 4.<br />
Since f is continuous on this interval, F 3 accepts all x for which 0 ≤ 2x ≤ 4.<br />
This implies that the domain for F 3<br />
is 0 ≤ x ≤ 2, and the graph that results is a<br />
horizontal compression of the graph of F. It might help for students to build a<br />
table to investigate the values F 3<br />
( 0), F 3<br />
( 1 2), F 3<br />
( 1), F 3<br />
( 3 2), and F 3<br />
( 2) .<br />
The graph of F 3<br />
on [0,2]<br />
(d) Students must next consider y = f (| t|) and must realize that it is an even<br />
function and that its domain requires 0 ≤ t ≤ 4 or, in other words, −4 ≤ t ≤ 4.<br />
The graph of y = f (| t|) is given below.<br />
The graph of f<br />
(| t|) on [–4,–4]<br />
52<br />
<strong>AP</strong>® <strong>Calculus</strong>: 2006–2007 Workshop Materials