+0  
 
0
904
7
avatar

1.) Find all values of x for which the line that is tangent to y = 3x - tanx is parallel to the line y - x = 2 

 

2.) Find the value of the constant A so that y = Asin3t satisfies the equation d^2y/dt^2 + 2y = 4sin3t 

 

3.) Suppose that f(x) is a peacewise function shown below, for what values of k is f?  

f(x) = (x^2) - 1, x <(or equal to) 1 

      =  k(x - 1), x > 1 

 

   a.) Is it continuous? 

   b.) Is it differentiable? 

 

 

Any help would be very much appreciated. :) 

Guest Mar 25, 2015

Best Answer 

 #7
avatar+26750 
+5

3.  

When x = 1 both x2 - 1 and k(x - 1) are zero, so the function is continuous for all values of k.

 

d(x2 - 1)/dx = 2x, which is 2 when x = 1

dk(x-1)/dx = k, which is 2 when k = 2, so the function is differentiable at x = 1 only when k = 2.

.

Alan  Mar 26, 2015
 #1
avatar+92805 
+5

1.) Find all values of x for which the line that is tangent to y = 3x - tanx is parallel to the line y - x = 2

ALWAYS remember that when you differentiate you are finding the gradient to the tangent of a curve

$$\\y=3x-tanx\\
y'=3-sec^2x\\\\
y-x=2\\
y=x+2\\
y'=1\\\\
1=3-sec^2x\\
-2=-sec^2x\\
2=sec^2x\\
\frac{1}{2}=cos^2x\\
cos x = \pm\frac{1}{\sqrt2}\\
x= n\pi\pm \frac{\pi}{4}\qquad n\in Z$$

1.) Find all values of x for which the line that is tangent to y = 3x - tanx is parallel to the line y - x = 2  - See more at: http://web2.0calc.com/questions/stuck-on-calculus#sthash.cD8MZGpc.dpuf
Melody  Mar 25, 2015
 #2
avatar+87301 
+5

I'm going to amend Melody's answer a bit

cos x  = ±1/√2    at   pi/4 ± n (pi/2)

 

  

CPhill  Mar 25, 2015
 #3
avatar+87301 
+5

2.) Find the value of the constant A so that y = Asin3t satisfies the equation d^2y/dt^2 + 2y = 4sin3t 

I assume this is

d2y/dt2 + 2y= 4sin3t

If y= Asin3t

dy/dt = 3Acos3t

And  

d2y/dt2  = -9Asin3t

So

d2y/dt2 + 2y= 4sin3t

-9Asin3t + 2(Asin3t) = 4sin3t

-7Asin3t = 4sin3t

A = -4/7

 

  

CPhill  Mar 25, 2015
 #4
avatar+92805 
+5

Chris maybe I am going a little balmy but I think your correction is identical to my original answer.

They are presented differently but I think that they are the same.  

Melody  Mar 25, 2015
 #5
avatar+87301 
+5

-------------------------------------------------------------------------------------------------

CPhill  Mar 25, 2015
 #6
avatar
0

Thanks guys, your explanations were really helpful. I'm however still stuck on the third one :/

Guest Mar 25, 2015
 #7
avatar+26750 
+5
Best Answer

3.  

When x = 1 both x2 - 1 and k(x - 1) are zero, so the function is continuous for all values of k.

 

d(x2 - 1)/dx = 2x, which is 2 when x = 1

dk(x-1)/dx = k, which is 2 when k = 2, so the function is differentiable at x = 1 only when k = 2.

.

Alan  Mar 26, 2015

12 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.