Stuck on Calculus

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.

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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

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

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

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

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

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.  

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