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AP Calculus

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Special Focus: The Fundamental<br />

Theorem of <strong>Calculus</strong><br />

The graph of y = erf ( x) is concave up on the interval ( −∞, 0 ) and concave<br />

down on the interval ( 0, ∞ ).<br />

2 0 2<br />

−t<br />

(e) erf ( 0)<br />

= ∫ e dt = 0.<br />

π<br />

0<br />

Using technology, here is a graph of y = erf ( x). Notice that this graph confirms<br />

the results obtained in parts (a) through (d).<br />

4. (a) h(x) is defined as the composition of two functions. Although we do not have a<br />

specific rule for the function f, we can still simplify the derivative of h:<br />

2 .<br />

h( x) = ( f g)( x) = f ( g( x)) = f ( x )<br />

We use the chain rule to obtain that h′ ( x ) d<br />

= = ′<br />

dx f ( x 2 ) f ( x 2<br />

)( 2 x)<br />

.<br />

(b) Let G( x) = x<br />

2 x −t<br />

10<br />

and F( x) = ∫ ( t − 4)<br />

e dt.<br />

0<br />

x<br />

Then ( F G)( x) = F( G( x)) = F( x 2 2<br />

−<br />

) = ( t − 4) e t 10 dt = H(<br />

x) .<br />

∫<br />

0<br />

2<br />

2 2 − 10<br />

As in part (a), H′ ( x) = ( 2x) F′ ( x ) = ( 2x)( x − 4)<br />

e x<br />

2<br />

2 −x<br />

10<br />

(c) H '( x) = 0⇔( 2x)( x − 4)<br />

e = 0 ⇔ 2x<br />

= 0 or x 2 − 4 = 0 or e x<br />

Thus we conclude that H′ ( x) = 0 ⇔ x = 0 or x = ± 2 (since e x<br />

never zero).<br />

.<br />

− 2 10<br />

=<br />

− 2 10 is<br />

0.<br />

122<br />

<strong>AP</strong>® <strong>Calculus</strong>: 2006–2007 Workshop Materials

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