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# Stuck on Calculus

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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

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. :)

Mar 25, 2015

#7
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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.

.

Mar 26, 2015

#1
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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

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

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
Mar 25, 2015
#2
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I'm going to amend Melody's answer a bit

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

Mar 25, 2015
#3
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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

Mar 25, 2015
#4
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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.

Mar 25, 2015
#5
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Mar 25, 2015
#6
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Thanks guys, your explanations were really helpful. I'm however still stuck on the third one :/

Mar 25, 2015
#7
+28029
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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