# SOLUTION: MATH 20A Columbia University Intervals and Tangent Line Exam

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Math 20A – Final exam

Problem #1

Part (a)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

lim f(x) = 1

x →0−

lim f(x) = 2

x →0+

lim f(x) = does not exist

x →0

lim f(x) ≠ lim+ f(x)

x →0−

x →0

lim f(x) = −2

x →3−

lim f(x) = −2

x →3+

lim f(x) = −2 see the music lower

x →3

Note that the condition of f(x) as x → 3 does exist… but is not correspondent to f(3) accordingly the

employment is discontinuous at x=3.

Part (b)

g’(x) = the appear of the tangent continuity at the sharp-end (x, g(x)). We distinguish the appear of that

tangent continuity = 6, we distinguish the sharp-end (x, g(X)) = (1, 30) so we can produce the tanget

continuity after a while that appear and sharp-end, and then use that tangent continuity to estimate g(1.5)

Tangent continuity

y = 6x + b

30 = 6(1) + b

B = 24

y = 6x + 24

estimated treasure of g(x) at x = 1.5

g(x) = 6*(1.5) + 24 = 33

Answer: g(1.5) = 33

Part (c)

Note that A(x) = complete of the employment plotted. The complete is the area lower the

incurvation plotted. So as hanker as the employment plotted is aloft the x-axis, the area is

increasing as x goes from left to just. While it’s lower the x-axis, it obtain be decreasing.

A is hollow up were the employment is increasing and hollow down where the employment is

decreasing. A reaches it apex treasure where the employment changes from increasing

area lower the incurvation to decreasing. That is, where the employment crosses the x axis.

(i)

(ii)

(iii)

A is increasing balance the interim [-2, 1]

A is hollow up balance the interims [-2 , -1) & (2 , 3]

A is hollow down balance the interim (-1, 2)

A is maximized at x = 1

Problem #2

Part (a)

Given

2sinx − (ln x)4 + 1

lim

x→∞ 7sinx + 3(ln x)4 − 3

Dividing through by x

2sinx (ln x)4 1

− x +x

lim x

x→∞ 7sinx

3(ln x)4 3

−x

x +

x

Distributing the condition

(ln x)...

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