So k is the graph (a)

Exercise 3

h is the only function that is defined on the whole line. Hence the graph is (d).k has oblique asymptote x since lim_x→±∞x/p|x⁴− 1| − x = 0. So k is the graph (a). The different between (b) and (c) is how flat the graph near x.

So we check derivative of f and g, the small one would correspond to graph (b) and the other to (c)(i.e. f is (c),g is (b)).

Exercise 6

Red area present f⁰⁰ > 0 and the pink area present f⁰ < 0,and the corre- sponding compliments keep blank.

Exercise 26

By page 247, the asymptotes of rational function, we have asymptotes y = 0, and x = 1, −2(since denominator has factor x − 1 and x + 2). Using product rule, we got y⁰ = (−x² − 2)/(x² + x − 2)² < 0. For convenience, we can express y as

y = 2/3

x + 2 + 1/3 x − 1. Then

y⁰⁰= 4/3

(x + 2)³ + 2/3 (x − 1)³.

This tell us y⁰⁰ > 0 as x > 1, y⁰⁰ < 0 as x < −2 and there exist reflection point amid (−2, 1).

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Exercise 32

Note that y = e^−xsin x actually is like sin x but decreasing its amplitude while x → 0. To show this, you can take derivative and you can see that y⁰ = e^−xsin (3π/4 + x) which has max-min points on 1/4π + nπ for n = 0, 1, ....

Taking second derivative, we have y⁰⁰ = −2e^−xcos x, which has reflection points on π/2 + nπ for n = 0, 1, .... Also y has intercepts as sin function on x-axis.

Exercise 36

y⁰ = (1 − 2 ln x)/x³, y⁰⁰ = (−5 + 6 ln x)/x⁴. We have maximum point on x = e^1/2. y⁰⁰< 0 as x < e^5/6, y⁰⁰ > 0 as x > e^5/6. Moreover, we have vertical asymptote y = 0 and horizontal asymptote x = 0.

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Exercise 39

y⁰ = 2

3x( 1 (x²− 1)^2/3) ,

y⁰⁰= [( 1

(x²− 1)^2/3)(−2

9 )(x²+ 3)]( 1 (x²− 1))

.This tell us y has minimum point on x = 0, y⁰⁰ > 0 on [−1, 1],y⁰⁰ < 0 on (−∞, −1) and (1, ∞). y⁰, y⁰⁰ is not defined on −1, 1.

Exercise 40

Let x = e^−h, then lim_x→0+x ln x = lim_h→∞^−h_eh = 0. For lim_x→0x ln |x| case, it is just by take x = −e^−h and do again.Hence we conclude that f (x) = x ln |x|

is well-defined continuous function on x = 0. Moreover, f is odd function. So only need to describe the graph for x > 0.f⁰ = 1 + ln x,f⁰⁰ = 1/x. Hence f is

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concave up and has minimum point on x = e⁻¹. Also f has intercept points with x-axis on x = 0, 1. f has no asymptote since f⁰ → ∞ while x → ∞.

Exercise 41

Since sin x is a bounded function, y go to zero as |x| → ∞. Hence there is asymptote y = 0. Moreover, since sin x osilates between 1, −1, it cross x-axis on nπ for n = 0, 1, −1, 2, −2, .... This exercise tell us even a function has asymptote , it might not lie on one side of asymptote when x large enough.

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