MA005: Calculus I

1. A–a, B–c, C–d, D–b

3. A–b, B–c, C–d, D–a

5. (a)–C, (b)–A, (c)–B

7. $f(x) = x^2 + 3$, $g(x) = \sqrt{x – 5}$, $h(x) = \dfrac{x}{x – 2}$
(a) $f(1) = 4$, $g(1)$ is undefined , $h(1) = –1$.
(b) Graphs of $f$, $g$ and $s$ are shown in Fig. 0.3P7.

(c) $f( 3x ) = ( 3x )^2 + 3 = 9x^2 + 3$, $g( 3x ) =\sqrt{3x – 5}$ (for $x ≥ 5/3$) $h( 3x ) = \dfrac{3x}{3x – 2}$
(d) $f( x+h ) = (x+h)^2 + 3 = x^2 + 2xh + h^2 + 3$, $g( x+h ) = \sqrt{x + h – 5}$ , $h(x + h) = \sqrt{x+h}{x+h – 2}$

9. (a) $m = 2$ (b) $m = 2x + 3 + h$. (c) $m = x + 4 (if x≠1)$ If $x = 1.3$, then $m = 5.3$. If $x = 1.1$, then $m = 5.1$. If $x = 1.002$, then $m = 5.002$.

11. $f(x) = x^2 – 2x$, $g(x) =\sqrt x$.
$m = \dfrac{f( a+h ) – f( a )}{h} = 2a + h – 2 (h≠0)$. If $a = 1$, then $m = h$. If $a = 2$, then $m = 2+h$.

If $a = 3$, then $m = 4 + h$. If $a = x$, then $m = 2x + h – 2$.

$m =\dfrac{g( a+h ) – g( a )}{h} = \dfrac{1}{\sqrt{a+h} +\sqrt a} = \dfrac{\sqrt{a+h} –\sqrt a}{h}$ . If $a = 1$, then $m = \dfrac{\sqrt{1+h} – 1}{h}$.

If $a = 2$, then $m =\dfrac{\sqrt{2+h} –\sqrt 2}{h}$ . If $a = 3$, then $m =\dfrac{\sqrt{3+h} – 3}{h}$.

If $a = x$, then $m =\dfrac{\sqrt{x+h} – x}{h}$.

13. (a) Approx. 250 miles, 375 miles. (b) Approx. 200 miles/hour.
(c) By flying along a circular arc about 375 miles from the airport.


15. (a) See Fig. 0.3P15.

(b) Max at $x = 2$. Min at $x = 4$.
(c) Largest at $x = 5$. Smallest at $x = 3$.


17. The path of the slide is a straight line tangent to the graph of the
path at the point of fall. See Fig.0.3P17.

19. (a) $s(1) = 2$, $s(3) = 4/3$. $s(4) = 5/4$.

      (b) $s(x) = \dfrac{x + 1}{x}$.

21.

23. On your own.