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avatar+1998 

1.

 

2.

 Jan 8, 2021
 #1
avatar+114361 
+2

(1).....really reaching back into my memory banks  here

 

I believe  that the answer is   y  = tan x +  C

 

Note  that the slopes shown in the  field  are always positive...and the slope  of the tangent is always positve (where defined).........also.....the other  cliue is that near 0,  the  slope of the tan  decreases   on  both sides of  0  and then increases dramatically   away from 0  to  pi/2  (and - pi/2)

 

Look at the graph here  :  https://www.desmos.com/calculator/uceau9xjx7

 

I hope that I haven't mis-lead you

 

cool cool cool

 Jan 8, 2021
 #2
avatar+114361 
+2

(2)  Again.....I'm not totally sure...but....

 

I think the answer is  y =  sec x

 

Look at the  graph here : https://www.desmos.com/calculator/v7cuzaovi5

 

As we approach  0  from the left side....the slope is  negative   and  on the other  side of  0 the slope is positive

 

Also....the other clue is that  on each side of the asymptotes on the secant, the slopes occur in "pairs"  negative-negative  and positive-positive....notice that this occurs in the slope field, as well

 

Hope this is correct   !!!

 

cool cool cool

 Jan 8, 2021
 #3
avatar+112034 
0

Questions like this are always confusing.

I think this is what it means.

** Sorry Chris, I only just realized that you had already given the same graph.

 

here is the graph of y=tan x

Can you see that the gradient (which is the derivative) is always positive?

As the graph approaches the asymptotes, the gradient becomes infinitely large.

In the middle of each section, the gradient appears to be about 1.  It is steeper than that everywhere else.

This looks the same as one of your graphs.

Which one is that?

 

 Jan 9, 2021
 #4
avatar+112034 
0

If you draw      y=cos x   or   y=sin x    you will see that the gradient of the curve is never infinite (there are no asymptotes) 

So neither of these could be sin x    or    cos x.

 Jan 9, 2021

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