curve sketching Q (1 Viewer)

[ToO LaZy ^*]

i)
sketch, without the use of calculus, on the same set of coordinate axes and over the domain -2pi <= x <= 2pi, the three functions:

f(x) = sinx , g(x) = sin²x and h(x) = f(x) + g(x) done

ii) on separate coordinate axes sketch y = Ln(h) cannot do

so how would you go about sketching this step-by-step?

 

[gman03]

whenever h(x) --> 0, then y -> -infinity
for h <= 0, y(x) is not defined
when h = 1, y = 0
when h = e, y = 1...
when when h is increasing, y is increasing.... and so on

Hope this help

 

[gman03]

oops i thought it was ln.... in that case i dunno about In

 

[ToO LaZy ^*]

Originally posted by gman03
oops i thought it was ln.... in that case i dunno about In

no sorry, i mean Ln (Lin)
i'll change it now

 

[:: ck ::]

umm if u know how to draw the first one

basically to draw the Ln graphs

when h(x) crosses y = 1

theres an x intecept

when h(x) hits the x axis
theres a vert asymptote

when h(x) is between 0 and 1
Ln graph is under x axis

thats all u really need to know

 

[ToO LaZy ^*]

Originally posted by :: ryan.cck ::
umm if u know how to draw the first one

basically to draw the Ln graphs

when h(x) crosses y = 1

theres an x intecept

when h(x) hits the x axis
theres a vert asymptote

when h(x) is between 0 and 1
Ln graph is under x axis

thats all u really need to know

hmm.. i don't really get it
h(x) never crosses y=1, it crosses at the origin

here's the answer..but i still don't know how to do it

 

[:: ck ::]

well look at ur graph

its always under the x axis coz as u said.. it never crosses y = 1

and as it crosses the origin u have all of these asymptotes .. which is present in the attachment u posted

 

[ToO LaZy ^*]

hmm..ok

 

[ToO LaZy ^*]

i got another q
sketch the curve
y = tan^-1(x) + tan^-1(1/x)
any takers?

 

[CM_Tutor]

Originally posted by ToO LaZy ^*
i got another q
sketch the curve
y = tan^-1(x) + tan^-1(1/x)
any takers?

Show that dy/dx = 0.
The domain is {x: x &ne; 0}
Evaluate tan^-1x + tan^-1(1 / x) at x = 1 and x = -1
You should be able to sketch it from there.

 

[ToO LaZy ^*]

ok got it

 

[schmeichung]

do u have the answer for this?
y = tan-1(x) + tan-1(1/x)

jus wanna check if im correct

 

[ToO LaZy ^*]

yeah..
it's a straight horizontal line from 0 -> infinity at y = pi/2
and another straight horizontal line from 0 -> -infinity at y = -pi/2

 

[CM_Tutor]

yeah..
it's a straight horizontal line from 0 -> infinity at y = pi/2
and another straight horizontal line from 0 -> -infinity at y = -pi/2

With open circles at (0, &pi; / 2) and at (0, -&pi; / 2).