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


see robust.

Math 20A – Final exam
Problem #1
Part (a)

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.


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)
2sinx − (ln x)4 + 1
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 +
Distributing the condition
(ln x)...

15 Million Students Helped!

Sign up to scene the ample answer

Source coalesce